Remember when away from math for a period of time, we lose our skills to solve problems quickly. So it's always a good idea to review the basics in the math class you are currently taking or review the most recent material. When school begins again, you'll be ready to go and not fall behind. Math is an ongoing process, what is learned previously needs to be mastered in order to do well in the material that is to come.

Suppose you have 100 inches of wire to form the skeleton of a rectangular box with a square base. What dimensions of the box will maximize the volume?

The formula for volume for the box is area of the base times the height. Side the base is square, the area of the base is x^2, with x being the length of a side of the square. Let's make the height "y", therefore the volume is (x^2)y.

The distance around the base is 4x since each side is of length "x", therefore it is 8x around the top and bottom of the box. The sides are each length "y", so the four sides have length 4y.

So we know that 8x + 4y = 100.

Now solve the equation for either x or y. Solving for y, we get y = (100 - 8x)/4 = 25 - 2x.

Substitute that in for the volume equation to get volume = x^2(25 - 2x) = 25x^2 - 2x^3. To maximize the volume, take the derivative of the volume equation, which equals 50x - 6x^2.

Set this equal to zero and solve for x to get, 2x(25 - 3x) = 0. Therefore x = 0 or 25/3. 

The dimensions of the box are 25/3 by 25/3 by 25/3, so the box is a cube.

A ratio that is seen many times in geometry, architecture and art is the "golden ratio", which equals approximately 1.618.

We can find the golden ration by dividing a line into two parts such that the longer part, which we can label "a" divided by the smaller part, labeled "b" is also equal to the entire length of the line "a + b" divided by the longer part "a".

a/b + (a + b)/a

It is said that the most appealing rectangle to the eye is in the golden ratio.

One such rectangle is one with length 10 and width 6.18.

a = 10
b = 6.18

10/6.18 = 16.18/10 which is approximately 1.618

Suppose you have a function f(x) = 4x^3 + 3x^2 + 5x + 2.  What is the slope function and how do can you check that your slope function is correct?

The derivative of f(x), f'(x) = 12x^2 + 6x + 5.

We can check by taking the derivative using the slope function at a point, say x = 1.  The derivative of that is 23.

Now use the mean value theorem, [f(b) - f(a)]/(b - a).  Use values for a and b very close to 0, one on either side to approximate the derivative at 0.

a =1.1
b =0.9

f(0.9) = 11.846
f(1.1) = 16.454

Therefore, we get (11.846 - 16.454)/(0.9 - 1.1)  = 23.04  so the slope function holds true.

Here's some simple rules to remember for logarithms.

log(base b) x = y can be rewritten in exponential form as b^y = x.

 Ex)  log (base 3)9 = 2 is the same as 3^2 = 9

ln x = y is the same as log (base e) x = y, which is e^y = x.

A natural log is basically a logarithm with base e, where e is approximately 2.718.

The log of a product is the same as the sum of logs.

log xyz = log x + log y + log z

The log of a quotient is the difference of the logs.

log x/y = log x - log y


Another useful rule is log x^2 = 2logx.  Basically the exponent can be moved in front of the log.

These basic rules will help anyone when trying to solve problems with logarithms.

Suppose a function  f(x) is differentiable and continuous on the interval [a,b]. If there exists a number c in the interval [a,b] such that f '(c) = 0, then the Mean Value Theorem applies.

For example..  suppose f(x) = x^2 + 3x. We want to test the Mean Value Theorem over the interval [-2, 1].

f'(x) = 2x + 3

Now set f'(x) = 0 and solve for x.

2x + 3 = 0, therefore x = -3/2.  Since -3/2 falls in the interval, the Mean Value Theorem applies.

An example where it doesn't apply. Suppose the interval is [-5,5]

f(x) = 2x..

f'(x) = 2

there is no value c where the derivative is 0 since the derivative everywhere along the function is 2. Therefore, the Mean Value Theorem does not apply.

Suppose you want to drain a swimming pool at the end of the summer and you know the pool drains at 10 cubic feet per minute. The pool is 100 feet long and 40 feet wide. For the first 50 feet, the pool was water depth of 4 ft. For the next 30 feet the water depth is 7 feet and for the final 20 feet the water depth is 10 feet. How long will it take the pool to drain?

We need to calculate the volume of the pool.  Recall the volume is length times width times depth.

For the first 50 feet the volume is 50(40)(5) = 10,000 cubic feet.  For the next 30 feet the volume is 30(40)(7) = 8,400 cubic feet. For the last 20 feet, the volume is 20(40)(10) = 8,000 cubic feet. Therefore the total volume of the pool is 26,400 cubic feet.

The time it takes the pool to drain is 26,400/10 = 2,640 minutes.

Wondering what everyone thinks about the common core math. I read that it's not necessarily meant to teach students to figure out an answer to a problem quicker, but it's more to teach children how to think. The rationale behind it is that kids don't need to know to do the calculations, per say, since a calculator can do all the work for them. The old traditional methods of adding, subtracting, multiplying and division are quicker, but many believe that kids just memorize the steps, the process by which to get the answer without even knowing why different things are performed. For example, why are numbers carried and borrowed?   As a long-time math tutor, I prefer the old traditional methods. I'd much rather use them and not allow kids to use calculators on the more simple math problems. Anyone have an opinion on the common core math methods?

When calculating the area under a curve over a certain interval, you can approximate it by getting the area of rectangles under the curve or equal widths. The more narrow the rectangle, the closer the approximation to the area.

The exact area can be found by integrating the function, which is the opposite of differentiation.

Suppose we want the area under the curve defined by the function f(x) = x^2 + 4x + 6 from [1, 4].

We integrate the function first. To do so we take 1/(exponent + 1)(coefficient and variable)^(exponent + 1). 

So x^2 becomes (1/3)x^3

4x becomes 2x^2

and 6 becomes 6x

Put it all together we have (1/3)x^3 + 2x^2 + 6x.  Now we substitute the values in the interval for x, first 4.

(1/3)(4^3) + 2(4)^2 + 6(4) = 77 1/3

Now substitute 1 for x to get

(1/3) + 4 + 6 = 10 1/3

Subtract 10 1/3 from 77 1/3 to get the final answer of 67.

Suppose the position of an object at time t is noted by the function s(t) = -4.9t^2 + 300.  The object is dropped from the top of a building. What is the average velocity of the object in the first three seconds after it is dropped?

The velocity, v(t), is found by taking the derivative, which is the rate of change of the object.

v(t) = -9.8t

Now we can used the mean value theorem, which is [f(b)- f(b)]/(b - a) for some interval a to b in which the function is differentiable. The interval is 0 to 3, so a = 0 and b = 3

f(3) = 255.9
f(0) = 300

(255.9 - 300)/(3 - 0) = -14.7 which is the average velocity. How can velocity be negative, one might think? The object is going downward, which makes it negative. Speed can only be positive but velocity can be both positive or negative depending on the direction.

Suppose f(x) = x^2 - 3x + 7, which is a parabola, and want to know the maximum and minimum on the interval [0,2]. 

First, we find the critical values, which is where the slope along the curve equals 0. To get that we take the derivative, set equal to 0 and solve for x.

f'(x) (derivative) = 2x - 3.

Set equal to 0 and we see that x = 3/2 or 1.5

Now, find the y coordinate of the endpoints of the interval and of the critical point.

f(0) = 7
f(2) = 5
f(1.5) = 4.75

Therefore the maximum on the interval is at 0,7 and the minimum on the interval is (1.5, 4.75).  In fact, since this is a parabola, the vertex is (1.5, 4.75), which is the minimum value of the curve.

suppose a car has a wheel with a radius of 15 inches. How many revolutions must the wheel make to cover a distance of 10 miles?

First you have to figure out what distance a wheel will cover in one revolution. That means we calculate the circumference of the wheel.

C= 2Pi(Radius)

C= 2(3.14)(15)

Therefore, the circumference is approximately 94.2 inches, which is 7.85 feet.

There are 52,800 feet in 10 miles. Therefore, there are 52800/7.85 = 6726 revolutions.

Suppose you have a conical tank that is 24 feet deep and 10 feet across. Water flows in at the rate of 10 cubic feet per minute. How fast is the depth rising when the depth of the tank is 8 feet?

We can use calculus to solve this problem.

The volume of a cone is 1/3(Pi)(Radius squared)(Height)

To solve this, use implicit differentiation. First we need to substitute a value for r in terms of h.  The we know that 5/12 = r/h, therefore r = (5/12)h.

V = (1/3)Pi(5/12 h)^2(h)
    = (1/3)Pi(25/144)h^3

Differentiate to get

V' = (1/3)Pi(75/144)h^2(h')

We know V' = 10, h = 8. Substitute those values into the equation and solve for h'.

10 =  (1/3)Pi(75/144)(8)h'

10 = 4.36h'

h' = 2.29

Therefore, the depth of the tank is rising at the rate of 2.29 cubic feet per minute.
 

Suppose you want to know the probability that you draw a jack or a red card from a standard 52-card deck.  You know there are four jacks in the deck, one of each of the four suits (hearts, diamonds, clubs and spades). There are 13 cards of each color, so there are 26 red cards. 

So one might think the probability is 4/52 for the jacks plus 26/52 for the red cards to get 30/52. But that isn't correct because if you do that you care basically counting the jack of hearts and jack of diamonds twice, since they are in both sets of outcomes.

To fix that, you have to subtract 2 from the total to get 28/52, simplified to 7/13.

The basic formula for two events that can both occur simultaneously, you take P(First event) + P(Second event) - P(intersection of the events).

Some students have difficulty determining the domain and range of a function. If you can graph the function using a graphing utility or making an x,y table and plotting points to create the graph, then seeing the domain and range is much simpler.

The domain is the set of x values, or the input of the function. The range is the output or the y-value of the function.

For example, if the function is f(x) = x^2 + 3x and the input is x = 1, the output is f(1), which is the same as y = (1)^2 + 3(1) = 4.

Looking at a graph, the amount the graph stretches horizontally is the domain and the amount the graph stretches vertically is the range.

If the function is just represented as an (x,y) table, the domain is all the x values and the range is all the y values.

Examples of using the chain rule, product rule and quotient rule for differentiation

f(x) = cos^3 (2x)

-3sin^2(2x)(2)

-6sin^2(2x)


f(x) = 2x(5x^2 + 3)^4

Use product rule and chain rule

2x(4(5x^2 + 3)^3)(10x)

80x^2(5x^2+3)

f(x) = 6x^2/(3x^2 + 4)^3

[(3x^2 + 4)^3(12x) - (6x^2)(3(3x^2 + 4)^2(6x))]/(3x^2 + 4)^6


Product rule is first term times derivative of the second plus second term times derivative of the first.

Quotient rule is denominator times derivative of numerator times numerator times derivative of denominator, all divided by denominator squared.

How do you find the inverse of a function? Here's some simple steps.

Suppose f(x) = 3x + 8

1. Substitute f(x) with y
2. Interchange y and x
3. Solve equation for y


In this example f(x) = 3x + 8 becomes y = 3x + 8.

Next interchange x and y to get x = 3y + 8.

Now solve equation for y to get (x  - 8)/3 = y

If you have two functions and you want to determine if they are inverses of each other, find the inverse of each. If the inverse matches the opposite function, then the functions are inverses.

For example,  f(x) = x^2 and g(x) = square root(x).  Get the inverse of f(x) first, which is

y = x^2

x = y^2

Solve for y to get y = square root(x) which is g(x).

Now get the inverse of g(x)

y = square root(x)

x = square root(y)

solve for y to get y = x^2, which is f(x). Therefore, f(x) and g(x) are inverses.

Here's a sample SAT problem that uses the concept of exponents.

2^x + 2^x + 2^x + 2^x  = 2^7.  What does x equal?

It's temping to say 7/4 because 2^x appears 4 times and 7/4 times 4 = 7. But that isn't correct.

You need to calculate 2^7 first, which is 128.

Now since 2^x is added to itself 4 times, you can rewrite the equation as

4(x^2) = 128

Therefore, dividing by 4 gives x^2 = 32. 

The value of x is five since 2 * 2 * 2 * 2 * 2 = 32

Suppose you have the graphs of f(x) and g(x) and want to find the derivative of f(x)g(x) at x = 2.  What you do is use the product rule for differentiation to get

f'(x)g(x) + f(x)g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively.

Then using the graph you can get the values of f(x) and g(x) at x = 2. To get the derivative at x = 2, you look at the slope of the graph at that point on the graph and substitute those values into the equation.

Sometimes students have problems remembering formulas for area or volume. One easy way to know the volume of a cylinder is to think of a cylinder as a bunch of very thin circles stacked on top of each other. Think of a CD and stacking them on top of each other to a certain height. Assuming there was no hole in the middle of the CD, it's basically a cylinder, but if you think of it as just a circle, you can get the area of the circle and multiply it by the height of the stack. That is the volume of a cylinder.

We know the area of a circle is Pi times radius squared or Pi(r^2). Therefore the volume of a cylinder is Pi times radius squared times height or Pi(r^2)h.

How do we determine whether or not an ordered pair is a solution to a given inequality? 

Suppose the inequality is 4x + 3y > 19 and the ordered pair is (2, 6)

Substitute 2 for x and 6 for y in the inequality.

4(2) + 3(6) > 19

8 + 18 > 19

26 > 19.

Since that is true, the ordered pair is a solution. If the statement was false, the ordered pair is not a solution.

If you have a graph of an inequality and want to know what portion of the graph is part of the solution, pick an ordered pair on either side of the line and substitute the value into the equation as done above. If the statement is true, that side of the line is shaded to represent the solution region. If if the statement is false, the other side of the line is shaded to represent the solution region.

Percent increase and decrease

Suppose the cost of a particular item last year was $320 and increased to $360 this year. What is the percent increase?

First, calculate the dollar amount of the increase, which is $40. From here we divide that number by the original amount of $320. Multiply that decimal result by 100 to convert to percent. So the increase is 12.5 %

Suppose a item depreciates in value by 20% every year.  The original value of the item is $1,200. What is the value of the item after two years?

We take 20% of $1,200 which is .20 times 1,200, which is 240. Subtract 240 from the initial 1,200 to get 960.  After the first year the value is $960.

Now take 20% of 960, which is 192. Subtract 192 from 960 to get 768. The value of the item after two years is $768.

Suppose two go-carts are driving around a circular track that is 360 feet in circumference. There is a camera in both cars that are 60 degrees apart. How are apart are the two cars?

The number of degrees around a circle is 360. So if the cameras are at a 60 degree angle from each other, that is 1/6 of the entire circle.

Since the circumference of the circle is 360 feet, we take 360 divided by 6 to get the distance between the cars. Therefore, the cars are 60 feet apart.

This method holds true no matter what the angle is between the cameras.

Here's a good sample SAT math problem.

Suppose a and b are real numbers and (a + b)^2 = 56 and a^2 + b^2 = 18. What is the value of  ab?

The key to the problem is knowing that when you expand (a + b)^2, you will have a^2 + b^2 and ab as part of the expansion, which makes the problem easy.

(a + b)^2 = a^2 + 2ab + b^2

Now substitute and solve for ab.

56 = 18  + 2

38 = 2ab

19 = ab

Suppose W,X,Y,Z are all digits with the following conditions:

X = W + Y + Z
W - 5 = Y
Z = W - 1

What is the 4-digit number XWYZ?

We can approach this by listing all the possible values starting with Y.
Y must be 0, 1, 2, 3, 4 since Y + 5 = W and the largest number that W or any of the numbers could be is 9
Therefore, we know that W has to be 5,6,7,8, or 9.

One we know that, it's easy to see that Z must be either 4,5,6,7, or 8 since it's 1 less than W.
X is the other letters added together. The only possible number X can be is 9, obtained when W = 5, Y =0 and Z = 4. Any other combination of those numbers will make X greater than 9.
The 4-digit number is 9504.

If 2^(abc) = 64, where a, b and c are positive integers, what is the value of a + b + c and what is 2^a = 2^b + 2^c?

The first thing we need to do is figure out what power we raise 2 to get 64.
2^2 = 4
2^3 = 8
2^4 = 16
2^5= 32
2^6 = 64

So we know abc = 6 and we know that a,b, and c are different positive integers. So a, b and c must be 1, 2 and 3 in any order since 1*2*3 = 6.

Therefore a + b + c = 6 as well and 2^a + 2^b + 2^c  = 2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14.

The problem is simply if one knows the rules for exponents and breaks the problem down into steps.

Here's some good math jokes, as nominated by teachers and Twitter.

Source:

http://news.tes.co.uk/b/news/2013/10/09/the-10-best-maths-jokes-as-nominated-by-teachers-and-twitter.aspx


@MaxineHowells2h
Talking sheepdog gets all the sheep in the pen for his farmer. He comes back and says ‘All 40 accounted for.’ Farmer says, ‘I’ve only got 36!’ Sheepdog replies, ‘I know, but I rounded them up.'

@AnnaMFortune
Two cats called '1,2,3' & 'un,deux,trois' had swim race across channel.
1,2,3 cat won because un, deux, trois cat sank!


Followed by seven brilliant runners-up:

@LocalSchoolsN22h
Hired an odd-job man to do 8 jobs for me. When I got back, he'd only done jobs 1,3,5, and 7

@TheRedshiftAcad3h
Have you heard about the mathematical plant? It has square roots

@ppelk1h
After careful investigation, it was found that aliens' heights were paranormally distributed

@sann06384h
I hit someone with a scientific calculator - I used the cosh button

@Mylosafe4h
What kind of tree could a maths teacher climb? = Geometry

@alexamcgregor4h
What do you get if you cross a maths teacher and a clock? Arithma-ticks!

@Maths_George
Last night I dreamed that I was weightless! I was like, 0mg

Suppose a rectangular field used to play soccer has an area of 7,800 square yards. Is the length is 70 yards longer than the width, what are the dimensions of the field?

Area of a rectangle is length times width. Let x = the width of the soccer field. That means the length of the field is x + 70.

Now take the formula for area to get A = x(x + 70)

We know area is 7,800 therefore,
7,800 = x(x + 60)
7,800 = x^2 + 70x
0 = x^2 + 70x - 7800

Factoring gives us
0 = (x + 130)(x - 60)
x = -130 or x = 60

Since we are talking about a length, x = -130 makes no sense, so the dimensions of the field are 60 by 130.

If you draw a straight line from home plate to first base and continue around the bases in such a fashion, you form a square with area of 8,100 square feet. What is the distance from home plate to second base rounded to the nearest tenth of a foot?

First you need to know that the area of a square is side times side. So we know side times side is 8,100. Since both sides are the same length, solve by taking square root of 8,100, which is 90.

Now the distance from home plate to second base is the length of the diagonal of a square. The formula for that is square root of 2 time the side. Multiply 90 by square root of 2 and you get approximately 127.3 feet.

Suppose you want to fence off two rectangular pieces of land that have the same area. You have 300 feet of fencing. The total length of both pieces of land are 2x feet (each x feet in length). The width of the land is y feet (2 y's on the outside representing the width of the overall and and y feet in width down the middle separating the two pieces of land)

What is the area of the land in terms of x?  What length and width will give the maximum area of the land?

You know that area of a rectangular is length times width.  So the overall area is x(y). But we want the entire area in terms of x.

Well we know the total amount of fencing is 300 feet, so x + x + y + y + y = 300. (just took all the dimensions and added them).

2x + 3y = 300

Solve for y to get 3y = 300 - 2x

and y = 100 - (2/3)x

So now we can get the area all in terms of x..

Area = x(100 - (2/3)x)

100x - (2/3)x^2

To find the value of x which maximizes area, you can use calculus or simply take -b/2a, where b is the number in front of the x and a is the number in front of the x^2.

So -b/2a = -100/[2(-2/3)] = 75

The length is 75. To find the length, substitute 75 in for x in the equation 2x + 3y = 300. Therefore, y = 50.

The width of the land is 50 feet for a total area of 3,750 square feet.

What is the equation of the line tangent line to the circle with center (3,4) at the point (6,0)?

The tangent line is the line the touches the circle and one point, in this case the line will touch at (6,0).  If you draw a segment from the center (3,4) to the point (6,0) the tangent line will be perpendicular to that segment. Perpendicular lines have slopes that are negative reciprocals of each other, so the slopes will multiply to equal -1.

So the slope of the segment from (3,4) to (6,0) is (0-4)/(6-3) = -4/3.  Therefore the slope of the line perpendicular is 3/4. 

3/4 times -4/3 = -1, which is what the slopes of perpendicular lines must multiply to.

Now take the equation y = mx + b where m is the slope and b is the y-intercept. Remember the y-intercept is where the line crosses the y-axis.

We know m = 3/4, we know x = 6 and y = 0.

So put the numbers in and solve for b.

0 = (3/4)(6) + b

0 = 4.5 + b

b = -4.5 or - 9/2

So the equation of the line is y = (3/4)x - (9/2)

Quick Guide to Finding the Greatest Common Factor

A factor is a number that divides evenly into another number. For example, 2 divides evenly into 12, so 2 is a factor of 12. Finding the factors is known as factoring. Suppose we want to find the largest number that is a factor or two or more numbers. That number is called the greatest common factor (GCF) of the numbers.
To find the greatest common factor, do the prime factorization of each number. Then find the lowest power of common factors and multiply them together. But what is meant by prime factorization and how is this done?
When each factor of a number is a prime number, the number is said to be in prime factored form, or prime factorization.

Examples:
Find the prime factored form (or prime factorization) of the following:
1. 24 Divide 24 by 2 to get 12. Then 12 can be divided by 2 to get 6. Next, 6 can be divided by 2 to get 3. That leaves us with 2 x 2 x 2 x 3. Therefore, the prime factorization is 2³ x 3.
2. 72
Divide 72 by 2 to get 36. Then 36 can be divided by 2 to get 18. Next, 18 can be divided by 2 to get 9. Finally, 9 can be divided by 3 to get 3. That leaves us with 2 x 2 x 2 x 3 x 3. Therefore, the prime factorization is 2³ x 3².
Notice how multiples of the same factor are written in exponential form. Note that if you are unsure what number divides evenly into another number, if the number is even it is divisible by 2. If the sum of the digits in the number is divisible by 3, then the entire number is divisible by 3. (18... 1 + 8 = 9, 9 is divisible by 3, so 18 is divisible by 3.
Now that we understand how to do prime factorization, we can find the greatest common factor.

Example:
Find the greatest common factor of 40 and 72.
40/5 = 8, so we have 5 x 8 as factors of 40. Notice that 8 can be factored into 2 x 2 x 2. Therefore, the prime factorization of 40 is 5 x 2 x 2 x 2. The prime factorization of 72 is 2 x 2 x 2 x 3 x 3 from the second example above. Notice the common factors are 2 x 2 x 2, so the greatest common factor is 8.

Example:
Find the greatest common factor of 45 and 135.
45/5 = 9, so we have 5 x 9 as factors of 45. Notice that 9 is 3 x 3. Therefore, the prime factorization of 45 is 5 x 3 x 3.
135/5 = 27, so we have 5 x 27 as factors of 135. Notice that 27 is 3 x 3 x 3. Therefore, the prime factorization of 135 is 5 x 3 x 3 x 3.
Notice the common factors are 3 and 5. The lowest power of each factor gives us 3 x 3 x 5. So, the greatest common factor is 45.

Example: Find the GCF of 6 a² b , 24 a² b² and 48 a³ b³ . Note that I am using * as the multiplication symbol instead of "x" for more clarity.
6a²b = 3 * 2 * a * a * b
24a²b² = 3 * 2 * 2 * 2 * a * a * b * b
48a³b³ = 3 * 2 * 2 * 2 * 2 * a * a * a * b * b * b
Notice the common factors are 3, 2, a, and b. The lowest power of each factor gives us 3 * 2 * a * a * b. Therefore, the greatest common factor is 6a²b.

Example:
Find the GCF of 15x²y³, 20y², and 45xy.
15x²y³ = 3 * 5 * x * x * y * y * y
20y² = 2 * 2 * 5 * y * y
45xy = 3 * 3 * 5 * x * y
Notice the common factors are 5 and y. The term 20y² has 2 as a factor but the other two don't, and 45xy and 15x²y³ have 3 and x as a factor, but 20y² doesn't. The lowest power of each factor is 5 and y, therefore the greatest common factor is 5y.
This quick guide with examples on prime factorization and finding the greatest common factor should assist any student having difficulty on these topics.
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With school underway next week or this week for some students, it's important to review material from the previous year. Math is one subject that builds upon previous material. Suffer in something previously and you'll probably struggle in the next class, particularly when it comes to algebra, even sometimes in statistics. We tend to use many of the basics in all math classes. So just be sure to review and if you run into trouble in your math class or have children that run into trouble, get a reliable tutor. Most schools will have a list of tutors. If not, you can find some online on various websites. Or you can consult me, as I've been tutoring for over 14 years now.

Try this little math puzzle.

Suppose you have a four digit number. The first digit is 1/5 of the last digit and the second and third digit together is 3 times the last digit. What is the number? (hint, the sum of the digits is 12)

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The answer it 1155.  The last digit must be a multiple of 5 since the first digit is 1/5. So the last digit must be 5 and the first digit must be 1.  The tricky part at first reading this is that the second and third digit being 3 times the last, you might think the second digit is 15 and the third is also 15, but that makes no sense, because a digit is a single number. the 2nd and third TOGETHER is 15, so 1 in the second digit and 5 in the third. So the answer is 1155.

Here's a simple, but fun activity for kids learning multiplication. Start at the red dot. Answer the multiplication problems in order. The answer will be one of the numbers connected by the black line. Move to that number. If you don't get any of the numbers listed, you are in correct. Keep going through the problems and eventually you will exit at one of the orange blocks. Good luck!

This shows young students how to subtract using borrowing. It's simple mathematics but good for children who are just learning how to subtract.

Here's a fun activity for kids just learning division. Fill in the question marks to complete the division problem.

Here's an example of long division, as will be shown in my kids book for math

Word problems are the most difficult for many students to master. What I try to do with all word problems is write an equation to the problem as I am reading it, if at all possible. Look for key words such as "sum" "more than" "difference" "times" "of" "is equal". Each one of them tells you a certain mathematical operation that goes in an equation

Difference  "minus"
More than, sum "plus"
Of "multiplication"
Is equal "equals sign"
Product "multiplication'
Quotient "division"
Square of "to the second power"


Look for these key words to help you set up your equations to your word problems more easily.

Here's a three by two multiplication problem that I illustrate in my new book that I plan to write. The idea is to provide easy step by step instructions to students just learning multiplcation.

I'm in the midst of writing a math book for kids. Here's a sample of how I explain a subtraction problem involving borrowing.

2 by 2 multiplication illustrated

No lattice multiplication here, this is simple, traditional, 2 by 2 multiplication explained with an illustration.

Suppose you had an arithmetic sequence such as 1, 4, 7, 10, 13, .....    and you wish to find the sum of the first 50 terms of the sequence. You could take all 50 terms and add them together, but that is time consuming. There are some simple formulas you can use to get the answer.

Sn (sum of the first n terms) is the average of the first and last term multiplied by the number of terms. Written as a formula this is Sn = n[(a1 + an)/2] where a1 is the first term and an is the last term.

To find the 50th term we use the formula

an = a1 + (n-1)d, where d is the common difference between the terms. You can clearly see that common difference is 3.

so the 50th term is 1 + (50-1)(3) = 148

Now to get the sum, we get Sn = 50(1 + 148)/2 = 25(149) = 3,725

Understanding Betting Odds and Potential Payouts

Although not exactly a math article, this does relate somewhat to math.

With football season approaching, many fans will start engaging in their fantasy football leagues and betting on games. Fans will check the numbers for point spreads, over/under and moneylines, to name a few. For the novice, this is all very confusing, not only to understand what the numbers mean, but also potential payouts and losses. I am not a gambler and am not taking a stand on whether or not someone should gamble, but have a strong statistical and overall mathematical background and can explain these betting concepts. So without any further disclaimers, call this "understanding betting odds 101."

The first popular betting line is called the "moneyline", which is a bet to who will win the game without a point spread or other variable. These lines are typically displayed first. When betting on the moneyline, the favorite will have a negative sign (-) in front of the number and the underdog will have a positive sign (+) next to the number. So suppose the Altanta Falcons are playing the Jacksonville Jaguars. Next to the Falcons you see -300 and next to the Jaguars you see +150. This means that you have to bet $3 on the Falcons for a profit of $1. If you bet $300 on the Falcons and the Falcons win, your payout if $400, for a profit of $100. So basically, it's a dollar profit for every three dollars you bet. If you bet on the Jaguars, you pocket $1.50 for every $1 you bet. Betting on strictly the winner of the game using the moneyline is typically called a "pick em" game.


The most common type of bet is the "point spread", which allows people to bet on the difference in score between the two teams. Once again, the team that is favorite will have a negative sign in front of the number, while the underdog has a positive sign next to the number. For example, suppose the Pittsburgh Steelers are playing the Dallas Cowboys and the Steelers are -6 on the point spread and the Cowboys are +6. That means the Steelers are a six point favorite, so if you bet on the Steelers using the point spread, they have to win by at least seven for you to win. If the Steelers win by six, there is a push and you get your money back. Any win of less than six or a loss and you lose the money you wagered.

Sometimes you will see a 1/2 or a .5 with the point spread. What is the significance of this? For example, you might see a game with the San Francisco 49ers against the Green Bay Packers and the Packers are -3.5 or -3 1/2. That means there is no chance for a push and some bettors may shy away from such a game. if you bet the Packers, they must win by at least four or you lose. There is no chance of getting your money back and breaking even when there is a 1/2 or .5 involved in the spread.

There is also the total score or the over/under, which lets you bet on whether or not the total score for both teams exceeds or fails to meet the predetermined amount. The over/under can also be for specific players total yards, touchdowns, teams combined yards, team wins for a season, etc. For example, suppose the Philadelphia Eagles are hosting the New York Giants and the over/under for the game is 48. If you bet "over", the teams have to score more then 48 points for you to win the bet. If they score exactly 48, there is a push and you get your money back. Again, if there is a 1/2 or .5, there is no chance for a push.

Typically, the moneyline given on the spread are -110 unless otherwise noted. If one side is receiving a lot of bets, the odds makers may adjust the lines accordingly in an effort to balance the action. Remember that you can also bet on a point spread on a whole game, part of the game, just look to see what kind of bets are being offered.

This guide is useful for those betting for money and even those who are picking games with their buddies without money involved, just for bragging rights. There are other more complex types of bets, but the one's listed above are the most common.

Enjoy the games, have fun and good luck with your bets!

I was reading a blog post by someone comparing an NFL quarterback from the 1960's and 1970's to one from this era, stating how the one back in the older era was overrated and worse in many ways than one from today.  A basis of this, was in his opinion, that the older QB had a completion percentage of 50.1 as compared to 55.7 of today's QB and the interceptions thrown were higher as well for the older QB than the current player.

But to accurately say that the older players was worse, you really need to compare his numbers to the other QB in the league, the average and standard deviation come into play. I know for a fact that completion percentage in the past era was much worse than the current era. Players may be better today offensively with the complex schemes, but I know the defense isn't allowed as much freedom in today's game. For whatever the reason, have to compare statistically.

Suppose the league average completion percentage in the 1960's and 1970's was 50 percent, therefore the older QB was right on the average.

But today, a 55 percent completion percentage may be way below a league average of, say 60 percent. So in essence the older QB did better compared to the rest of the league even though his actual percentage was lower.

The same holds true for other statistical categories.

Solving Trigonometric Equations

A trigonometric equation is an equation which has a trigonometric expression involving a variable, such as sinx, cosx, tanx, cotx, secx, cscx. There are the six trigonometric functions I mentioned in a previous introductory article on trigonometry. Some trigonometric equations are true for all values of the variable, such as 1 + tan²x = sec²x, 1 + cot²x = csc²x and sin²x + cos²x = 1. But many trigonometric equations are defined for only some values of the variable.

Fox example, suppose the equation is sinx = √2/2. We know from the unit circle, or we can solving using a calculator, that 45 degrees is a solution for x because sin 45 = √2/2. This is not the only value that satisfies the equation. Sine is positive in the first and second quadrant, which can be seen on the sine curve. So the answer is 45 degrees and the 45 degree angle in the second quadrant, which is 135 degrees. If answering in radians, it's π/4 and 3π/4.

Let's try an equation involving a single trigonometric function such as 3cosx - 2 = 5cosx - 1. We solve such equations the same as solving a basic algebra equation involving a single variable. We isolate the function to one side of the equals sign and solve for the variable. In this case, we want to get cosx on the left side of the equation and everything else to the right side of the equals sign. Therefore, we can subtract 5cosx from both sides. That gives us -2cosx - 2 = -1. Notice we have -2 on the left side of the equation. But we all we want is cosx on that side, so add 2 to both sides. Doing that gives us -2cosx = 1. From here we simply divide both sides by -2 to get cosx = -1/2. From our knowledge of the unit circle we know it's a 60 degree angle, but since the value is negative, it's not in the first quadrant. Cosine is negative in the second and third quadrants. So the answers are 120 and 240 degrees. In radians, the answer is 2π/3 and 4π/3. Note that we convert degrees to radians by multiplying each degree measure by π/180.

Suppose the equation is in quadratic form, meaning in the form au² + bu + c = 0. where u is a trigonometric function and a, b and c are real numbers and a cannot equal zero. We generally can solve this kind of equation by factoring, which I explained in a previous article.

Let's try the equation 2sin²x + sinx - 1 = 0. We'll attempt to solve this by factoring. Remember we use two sets of parentheses, each with two terms separated by an + or - sign. The first term in each must multiply to 2sin2x, the last terms must multiply to -1 and when multiplied out must give you sinx in the middle. Putting it together we get (2sinx - 1)(sinx + 1) = 0. We know this is correct because multiplying this out we get 2sin²x + 2sinx - sinx - 1, which is 2sinx²x + sinx - 1, the original expression. We set each factor equal to zero and solve. Therefore we have 2sinx - 1 = 0 and sinx + 1 = 0. Solving for x in the same manner we did in the previous example, we get x = 30 and 150 degrees and x = 270 degrees. In radians, the answers are π/6, 5π/6 and 3π/2.

You may also encounter a problem which involved multiple angles. An example of such an equation is cot3x = 1. To solve this equation, we are still looking for the values of x from 0 to 360 degrees (one rotation around the unit circle). We know cot is 1 at 45 degrees and 315 degrees, in radians this is π/4 and 7π/4. Therefore 3x =π/4 and 7π/4, we will add 2π to each to get 9π/4 and 15π/4 and again to complete three rotations around the unit circle. This gives us 17π/4 and 23π/4. If the problem was cot4x we'd need four rotations and so on. Now 3x is equal to all of those values. So in order to solve for x, we divide each angle to 3 to get π/12, 7π/12, 3π/4, 5π/3, 17π/12 and 23π/12.

There are more types of trigonometric equations, such as those using identities to solve, which I'll deal with in another article. Hope this guide will be useful for those that need to learn how to solve basic trigonometric equations.

Discontinuity

In some cases, a function will have discontinuity, which is a value for x where the function is undefined. It could be removable discontinuity, in which case there is a "hole" in the graph, or non-removable discontinuity, which in many cases will be an asymptote.

Take for example the problem  f(x) = (x^2 - 1)/(x - 1).  This is undefined for x = 1.

Factored this becomes (x - 1)(x + 1)/(x - 1).  When simplifying we are in essence "removing" the x -1 from the numerator and the denominator and are left with x + 1.

The graph of x + 1 is a line with slope 1 and y-intercept of 1. But since we know x is undefined at 1, we have a hole at the coordinate (1,2) represented by an open circle.

Here's another example which has some discontinuity.

f(x) = (x^2 + 6x + 8)/(x^2 -2x - 8)

When factoring we get (x + 4)(x + 2)/[(x -4 )(x + 2)]

This function is undefined at x = 4 and x = -2.  Notice the removable discontinuity at x = -2. The other discontinuity at x = 4 is non-removable.

Which math should I take in college?

If you are a math major or statistics major, the majority of your courses will be laid out for you with the exception of major electives, in which you'll have a variety of math courses to choose from. Take all the statistics and probability courses you can if you are a statistics major. 

If you are going into a science based field of study or any type of engineering major, take calculus your freshman year.  Make sure you take a course in complex variables, which has engineering applications.


If you are not going into a math, science, statistics, or engineering based field,  your college may offer college algebra. If not, take an introductory statistics course. You may only be required to take a year's worth of math.

If you are in a business related field, take math that relates to finance or accounting principles.

Those who have studied a course in statistics are familiar with a histogram or frequency distribution. In statistical process control or SPC, the Pareto chart is a frequency distribution or histogram of attribute data arranged by categories. For example, suppose we consider defects on an submarine used in a naval application. We can plot the total frequency of occurrence of each type of defect on the y-axis against various types of defects on the x-axis. With this type of chart a user can quickly see which types of defects occur most frequently and less frequently. The cause of the defects that occur most frequently should probably be identified and addressed first.

 

One must notice that the Pareto chart doesn't list the defect in order of importance, rather just in order of frequency. For example, a certain defect may only occur three percent of the time, but if it's one that could end up in scrapping the submarine. So, in some cases, the defect occurring most frequently may not be the one to take care of first. How do we determine what to take care of first? There are two methods, first you could use a weighted average to modify the frequency counts, meaning that we emphasize the defects that are more significant. You could also have a second chart to consult that lists that has a large cost exposure.

There are variations to the basic Pareto chart I discussed above. Suppose we have components on a circuit board labeled by part number and assume that we know the percentage of defective components. We could have a chart that lists the percentage of defects on the vertical axis and the component number on the horizontal axis.

Suppose we know the components come from three suppliers, A, B and C. We could have a chart with the number of defective components on the vertical axis and the component number listed by supplier on the horizontal axis. For example, suppose for component number 1 that there are 30 overall defects, 20 from supplier A, 8 from supplier B and only 2 from supplier C. The vertical bar above component 1 would extend to 20 (labeled A), from 20 to 28 (labeled B) and from 29 to 30 (labeled C). A chart of this sort quickly enables a used to see which supplier provides a disproportionally large amount of defects.

Another variation that is extremely useful is one that has two scales on the y-axis. On the left side is the number of defects and the right side is total percentage of defects. Suppose there are defects of types A, B, C and D. There are 100 defects total, 35 of which are type A, 30 are type B, 25 are type C and 10 are type D. We have a bar that extends vertically above A up to 35, a bar vertically over B up to 30 and so on. Now we place dots which match the cumulative percent on the right side. For example, 35 is 35 percent of the total defects, so place a dot above defect A which lines up with 35 on the right side. Now B is 30, but since we are doing cumulative, we add it to 35 to get 65 and place a dot that lines up with 65 above B. With C, the dot will go at 90 since (65 + 25 = 90) and the final dot goes above D at 100. Connect the dots to complete the chart.

Generally speaking, the Pareto chart is a very useful SPC problem solving tool. There are five others tools which I may discuss in future articles on statistical process control.

The derivative can be thought of as the slope of a curve at any given point.  You can calculate the derivative using the definition, which is lim (h approaches 0) [f(x + h) - f(x)]/h.  Note that there are formulas to calculate the derivative which is much simpler, but will use the definition here just so you see how this works.

Suppose f(x) = x^2 + 5x

f(x + h) = (x + h)^2 + 5(x + h)

lim (h approaches 0) [(x + h)^2 + 5(x + h) - (x^2 + 5x)]/h

                              = (x^2 + 2xh + h^2 + 5x +5h - x^2 - 5x)/h

                              = (2xh + h^2 +5h)/h

                             = 2x + h + 5

                             (put in zero for h to get)    2x + 5

In a previous article, I showed how a control chart is a major tool in statistical process control. A control chart with the center line being the sample average to monitor the process mean, upper control limit and lower control limit, it's important to know the sample size that will give the size of the shift we wish to detect in the process. It's also very important to know the frequency in which we wish to sample.

Suppose we have a process where we are measuring the average ring diameter over many samples. Let's say the sample average which is the center line is 64 mm and the probability of detecting a shift from 64 mm to 64.02 mm increases as the sample size increases. If the process is large, we use smaller sample sizes than those if the shift of interest is relatively small. In general, larger samples will make it easier to detect small shifts in the process.For example, maybe a sample size of 15 will detect a shift of .01 thirty percent of the time. A sample size of 10 may only detect a shift of .01 twenty percent of the time.

As for the frequency of sampling, the most desirable situation is to take large sample sizes very frequently to best detect even the smallest shifts in the process. But that is usually not possible economically. What is general done is small sample sizes can be taken at short intervals, say sample of 10 every 15 minutes. Or take larger sample sizes at larger intervals, say sample of size 50 every hour.

Another way to handle sample size and frequency is through what is known as the "average run length" of a control chart, or ARL. It's calculation is simple, just 1 divided by p, where p is the probability that any point is out of control (outside of the upper control limit or lower control limit).

For example, if a control chart has limits of 3 standard deviations from the mean (center line), then ARL = 1/0.0027 = 370. Note that .0027 is found from a cumulative standard normal distribution chart. Many calculators will calculate this as well.

What does the ARL mean? It means that the average run length of any process in control is 370. In other words, even if the process is in control, an out of control signal will occur every 370 samples, on average.
There are other factors to be taken into account when trying to more accurately answer the questions of sampling frequency. Such factors include the cost of sampling, probabilities of various types of shifts in the process to occur, losses associated with an out of control process and more.

What do you think of this? Multiplication by drawing lines? hmmmm!!

http://sfglobe.com/?id=770&src=share_fb_new_770

Basics of a control chart

The best way for a product to meet a customer's fitness for use, it should be produced in a consistent and reliable manner. It must be capable of operating with very little change or variability around the optimal dimensions of the quality characteristics of the product. There are may problem solving tools helpful in achieving stability and reducing variability. One such tool is the control chart. What are the basic principles of the control chart?

 

A control chart is a graphical display of a quality characteristic that has been computed or measured from a sample versus a sample number or time. Think of the chart as an x-axis and y-axis, focusing only on the first quadrant values (zero and positive values only). The x-axis would have time or sample number, while the y-axis is the sample quality characteristic. Examples could be "average ring diameter", "copper concentration", or "sample fraction nonconforming". The chart contains a horizontal center line which represents the average value of the quality characteristic measure when the system is in the "in-control" state.

Two other horizontal lines, called the "upper control limit" (UCL) and "lower control limit" (LCL), are shown.
Sample values are plotted according to the measure of the quality characteristic. A process is said to be in control when all the points plot inside the control limits, in which case no action is required. If a point plots outside the control limits, the process is out of control and quality control experts will investigate to determine the cause of the out of control process and what course of action to take to get the process back in control.
Even if all the points plot inside the control limit, the process could be out of control if the points are not in a nonrandom manner. For example, if 17 of 20 sample points are below the center line but above the LCL and only 3 are above the center line and below the UCL, we would suspect that something is wrong. The points plotted should represent a random pattern.

Let's examine the statistical basis of the control chart. Suppose we have an x-bar control chart for piston-ring diameter. Note that x-bar is the sample average. Suppose the average ring diameter is 70 mm and the standard deviation of the process is 0.015 mm. If sample of size n = 10 were are taken, the sample standard deviation of x-bar is the standard deviation of the process divided by the square root of n. Therefore, the standard deviation of x-bar is 0.0047. If we use the 3-standard deviation control limits, the UCL would be 70 + 3(.0047) = 70.0141 and the LCL would be 70 - 3(.0047) = 69.9859. Now we can plot the sample values and see if they fall within the control limits and if they display a random pattern.

The most important use of a control chart is to improve the process. In real world applications, most processes do not operate in statistical control. Therefore, we will look for assignable causes for this out of control process. If they can be eliminated, the variability will be reduced and the process will be improved, hopefully to the point of being back in control.

This guide gives the very basics of what control charts are and how they are used. In upcoming articles, I will examine how to choose control limits, how to analyze patterns on control charts, and the other six tools of statistical process control.

Consistent practice is the key

Remember that math is unlike many other subjects where you don't have to necessarily remember a topic when moving on to another topic. But in math, everything is a building block to the next topic. That's why it's important to build a solid foundation, learn the basics of mathematics like the back of your hand. Practice is the key to get better at simple addition, subtraction, multiplication and division. When you progress into the pre-algebra, algebra, geometry and higher mathematics you have to be able to master each or you will fall behind.  Buy a good math book and work on the side. Many times I will work with students over the summer so they don't get rusty or forget what they learned the previous school year. Practice, practice and more practice is the key to keep your math skills sharp.

Understanding Parametric Equations

Suppose you wish to track the flight of a ball thrown and know how high the ball is after the ball have traveled 150 feet horizontally if the ball flight is represented by a parabolic curve. The curve can be represented as a function of time using two equations, one related to x (the horizontal distance) and t (representing time) and one related to y (the vertical distance) and t.

For example, suppose a player throws a ball from a height of six feet at an initial speed of100 feet per second at a 45 degree angle to the horizontal axis. After t seconds the location of the ball can be described as x = 100cos(45)t and y = 6 + 100sin(45) - 16t². The x-component is the ball's horizontal distance in feet and y is the ball's vertical distance in feet. Where is the ball located 1.5 seconds after it is thrown? We substitute 1.5 in for t in both equations to get x to be approximately 106 feet and y to be approximately 40.7 feet. We can graph the movement of the ball by substituting values for t and plotting the values of x and y as the ordered pair (x,y) and connecting the dots.

Suppose we want to eliminate the parameter. We do this by solving for t in one of the two equations and substituting the value in for the other equation. For example, suppose we have the parametric equations x = 2t and y = t² + 2. We'll solve for t in the first equation since it's easier. Therefore dividing both sides of the equation by 2, we get t = x/2. Now using t = x/2 and y = t² + 2, we get y = (x/2)² + 2. Notice we just did a simple substitution, putting x/2 in for y in for t in the second equation. Simplifying the equation, we get y = (1/4)x² + 2. One familiar with the equations of conic sections knows that this is a parabola. Written in standard form the equation is x² = 4(y - 2). This is a parabola that opens up with vertex at (0, 2).

Suppose we want to graph a curve represented by the parametric equations x = t + 1 and y = √t. First use y = √t and square both sides to solve for t. Therefore we get y² = t. Now substitute y² for t in the first equation to get x = y² + 1. We Know this is a parabola with the vertex at (1,0) and opening to the right. You may be tempted to graph the entire parabola, but remember the restriction on t in the equation y = √t. This is only defined then t is greater than or equal to 0. Therefore, y cannot be negative and we only graph the lower half of the parabola.

Let's try one more example. Suppose a curve is represented by the parametric equations x = 3sin t and y = 5cos t. First get the trigonometric function by itself by dividing the first equation by 3 and the second equation by 5. Doing so, gives us x/3 = sin t and y/5 = cos t. Now square and add these two quantities and you'll see this all fall into place. Squaring and adding gives x²/9 + y²/25 = sin²t + cos²t. Recall from the Pythagorean identities that the right side of this equation equals 1, therefore we have x²/9 + y²/25 = 1.

Those familiar with the equations of conic sections will see immediately that this is an ellipse centered at (0,0) with y being the major axis of length 10 and x being the minor axis of length 6. The ellipse will be elongated vertically 10 units and horizontally 6 units.

Parametric equations are very useful in solving problems involving vertical and horizontal distance. This guide should give students the basics on understanding and using parametric equations.

The six functions used in trigonometry are sine, cosine, tangent, cotangent, secant and cosecant. But there are thousands of relationships that exist among these functions. Proving a trigonometric relationship requires imaginative problem solving skills and algebraic factoring. For those that enjoy mathematics, as well as solving puzzles, you'll feel a sense of accomplishment verifying identities.

What are the fundamental trigonometric identities? The reciprocal identities are the facts that the reciprocal of sine, cosine and tangent are the three trigonometric functions cosecant, secant and cotangent, respectively. Therefore, it holds true that the reciprocal of cosecant, secant and cotangent are sine, cosine and tangent.
Since sine is opposite divided by hypotenuse and cosine is adjacent divided by hypotenuse and tangent is opposite divided by adjacent, it holds true that tangent is sine divided by cosine. Therefore, cotangent is cosine divided by sine.

Recall the Pythagorean Theorem which states that the sum of the square of the legs of a right triangle equals the hypotenuse squared (a² + b² = c²). From this relationship we know that sin²x + cos²x = 1, 1 + tan²x = sec²x and 1 + cot²x = csc²x. The pythagorean identities can be proven, but it's more important to know the identities, as they are used frequently in verifying other identities.

Let's verify some identities by changing to sines and cosines. That is often a method used to verify identities. Always work from the side of the equation that is most complicated, proving it equals the side of the equation that is less complex.

For example: Verify that (cscx)(tanx) = secx.
Change every function in terms of sine and cosine. Therefore we have
(1/sinx)(sinx/cosx) = secx
Notice the sinx in the denominator and the sinx in the numerator cancel each other, so we are left with 1/cosx and we know from the reciprocal identities that it equals secx, so we have verified the identity.

Let's try a more complex example.

Verify that (cosx)(cotx) + sinx = cscx
Again, work with the left side of the equation since it's the more complex side and change everything in terms of sine and cosine. Therefore, we get
cosx(cosx/sinx) + six = cscx
cos²x/sinx + sinx = cscx
Getting a common denominator of sin²x, the equation becomes
(cos²x + sin²x)/sinx = cscx
Notice that cos²x + sin²x = 1, therefore the equation becomes 1/sinx which we know from the reciprocal identities is cscx. The identity is verified!


Let's try one more example. This is the most difficult of the three.


Verify that sinx/(1 + cosx) = (1- cosx)/sinx
The method used in the previous example does not apply here, since everything is already in terms of sine and cosine. So now we have to get creative and find out how we can use a pythagorean identity to solve this. The pythagorean identities have squared terms in them, in particular in this case sin²x + cos²x = 1. Notice we can maneuver the equation to get 1 - sin²x = cos²x and 1 - cos²x = sin²x. If we work on the left side of the equation and multiply the numerator and denominator by 1 + cosx, we will get a desired result.
sinx/(1 + cosx) * (1 - cosx)/(1 - cosx) = sinx(1 - cosx)/(1 - cos²x)
Notice the 1 - cos²x in the denominator, which is perfect as it matches up with the manipulated version of the pythagorean identity above. The value of 1 - cos²x = sin²x. The equation now becomes sinx(1 - cosx)/sin²x. The sinx in the numerator cancels with a sinx in the denominator, leaving us with (1 - cosx)/sinx. The identity is now verified!

There are many examples of verifying identities in any trigonometry or pre-calculus book. Learn the identities and strategies for manipulating equations. Practice and you'll become a master and solving such problems

Here's an interesting article about improving your mental math skills and how it can benefit you in getting a consulting job. But even if you aren't going for a consulting job, honing your mental math skills can benefit you in every day life.

http://www.consultingfact.com/blog/how-to-improve-your-mental-math-for-the-management-consulting-application/

During my nearly thirteen years of tutoring math, there has been confusion among students when it comes to prime numbers. Part of the reason is the definition of prime numbers seen in most textbooks, which is that a prime number is any natural number whose only factors are 1 and itself. Using that definition, almost everyone will say that 1 is a prime number, when in fact it is not.

 

The true definition is that a prime number is any natural number that has only two factors, which generally is 1 and itself. That definition excludes 1, which only has one factor, because the only numbers that multiply to give you 1 is 1 and 1. All other natural numbers that are not prime are called composite numbers.
I have a way of teaching students the prime numbers from 1 to 100 using a few rules and tricks which make the explanation a lot clearer. You can start eliminating numbers as possible prime numbers based on the following:

*Any number that ends in 0 is divisible by 10 and 5.
*Any number ending in 5 is divisible by 5.
*Any even number is divisible by 2.
*Any number whose digits added is divisible by 3.
The entire number is divisible by 3. An example of this is 39. 3+9 = 12, which is divisible by 3, so 39 is also divisible by 3.
*Any 2 digit number where both digits are the same is divisible by 11.
*Also any number that has a square that is a whole number is not prime.

By applying these rules you can eliminate the following as possible prime numbers:
4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,
50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,88,90,
92,93,94,95,96,98,99, and 100.

The numbers remaining (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,87,89,91,97) are prime.

What if one wants to know the prime numbers greater than 100? There are a few good methods to read and study. The Sieve of Eratosthenes is an ancient method to find any prime number up to a specified number. For a faster, although more complex method to find prime numbers up to a specified number, one can use the Sieve of Atkin. The Sieve of Atkin is an optimized version of the Sieve of Eratosthenes and involves dividing numbers by 60, getting the remainder, and using the remainder to solve more complex quadratic equations. Most of the time a computer or calculator would be needed to solve such equations.

The rules I give are enough to determine prime numbers between 1 and 100, which is generally enough for all practical purposes. I hope this information will be useful for any student to find the prime numbers between 1 and 100.

Trigonometry, measurement of triangles, is used in various occupations including engineering, architecture and navigation. We start with the study of trigonometry by examining six functions. These trigonometric functions are determined by inputs that are he measures of acute angles of a right triangle. The outputs are the ratios of the length of the sides of right triangles. 

 

If you draw a right triangle, that is a triangle with one right (90 degree) angle and two acute angles, note a base angle as Ө. We note the sides of the triangle based off this angle and the right angle. The side opposite the right angle, which is the longest side of a right triangle is the hypotenuse. The other sides are opposite of Ө and adjacent to Ө. It's important to know the distinction between the sides because the six trigonometric functions are based on the ratio between these sides.

The trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. If you have a scientific calculator, you may have seen buttons with sin, cos and tan on them. Those are the trigonometric functions sine, cosine and tangent. Sine is the length of the side opposite Ө divided by the length of the hypotenuse. Cosine is the length of the side adjacent to Ө divided by the length of the hypotenuse. Tangent is the length of the side opposite of Ө divided by the length of the side adjacent to Ө. To figure out cotangent, secant and cosecant, they are simply the reciprocals of tangent, cosine and sine, respectively.

An important thing to note is that the value of the trigonometric functions depend only of the size of the angle Ө and not on the length of the sides. For example, a triangle with opposite side 2 and adjacent side 1 has the same trigonometric functions as a triangle with opposite side 30 and adjacent side 15, since the ratio between the two is 2:1 in both cases.

Now that we know the basic trigonometric functions, let's evaluate them in an example. Suppose we have a right triangle with the length of the opposite side of angle Ө to be 4 and the length of the adjacent side to angle Ө to be 3. We can find the values of the six trigonometric functions once we know the length of the hypotenuse. We can get that by using the Pythagorean Theorem, which is "a squared" plus "b squared" equals "c squared", if a, b and c are the sides of the triangle, c being the hypotenuse. Using the equation, we get c = 5. Therefore sine is opposite divided by hypotenuse, which is 4/5. Cosine is adjacent divided by hypotenuse, which is 3/5 and tangent is opposite over adjacent, which is 4/3. Cotangent is the reciprocal of tangent, so cotangent is 3/4. Secant is reciprocal of cosine, so secant is 5/3. Cosecant is reciprocal of sine, so cosecant is 5/4.


We can also find the angle measures of a right triangle using trigonometric functions. If you've ever noticed sin^-1 , cos^-1 , and tan^-1 on your calculator those are the inverse sine, inverse cosine and inverse tangent buttons and are used to find the angle given a the trigonometric ratio. For example, in our above problem we know sine = 4/5, cosine = 3/5 and tangent = 4/3. Knowing these ratios we can get the value of Ө and therefore the other angles of the triangle. Take inverse sine of 4/5 and we get 53.1 degrees, which will be the same as taking inverse cosine of 3/5 and inverse tangent of 4/3. All will give the value of Ө at 53.1 degrees. Since we have a right triangle, we know one angle is 90 degrees and since the angles of a triangle add to 180 degrees, the third angle is 36.9 degrees.

These are just some of the basics with right triangle trigonometry. There is much more to learn, but understanding these basics are key to progressing to more difficult topics and applications of trigonometry.

When adding fractions with unlike denominators, we must change each fraction so they have the same denominator. Here's a simple example to start with. 

1/2 + 1/4

To add these two fractions, we must get a denominator that is the same. This is called the common denominator. We can get a common denominator by multiplying the two denominators, in this problem it would be 2 x 4 = 8. But that is not the lowest common denominator, which is preferred when adding and subtracting fractions. To get the lowest common denominator, let's look at all the multiples of 2 and then the multiples of 4.

Multiples of 2 are

2, 4, 6, 8, 10, 12, …...

Multiples of 4 are

4, 8, 12, 16, 20, …..

The lowest common denominator is the smallest number that is the same in both sets of multiples. Notice the 4 is bold in each set. That is the lowest common denominator.

Now that we have the common denominator, we have to make ½ into an equivalent fraction with 4 as the denominator. . If we multiply the numerator and denominator by 2, we get 2/4.. Notice ¼ already has a denominator of 4, so we don't have to change this fraction in order to add. We now get...1/4 + 2/4 = 3/4

Check out this site for many interesting photos and tidbits of information about math and science. Here's one from the page of interest. Enjoy!

 http://www.sciencedump.com/subcategory/math

To my mathematically minded friends...this book sounds interesting. Check it out and let me know what you think. I will have to read this and I'm sure much of it will be fascinating!


 http://www.brainpickings.org/index.php/2012/04/16/in-pursuit-of-the-unknown-ian-stewart/?utm_content=bufferacb2c&utm_medium=social&utm_source=twitter.com&utm_campaign=buffer

Over the course of many years, I have come across mathematical puzzles and tricks that can baffle even the best mathematical minds. But what most of these tricks come down to is simple algebra. Here's some math tricks that you may have encountered. but couldn't figure out. Enjoy the art behind the mathematical magic presented in the next several paragraphs.

Here's some mathematical "magic" you can perform with a single person or many people in an audience. Pick a person and ask that person to pick any single or double digit number. Then ask the person to do the following, double the number, add 14, divide by two and subtract the original number. You will astound most people when you ask, "Is the number you are now thinking of seven?" Try this yourself and you'll see that the answer will always be 7.

There reason is the simple algebra behind this. The number chosen is x, double the number to give you 2x. Add 14 to get 2x + 14. Divide by 2 to get x + 7. Now subtract the original number x, to get x + 7 - x = 7. This will work no matter what number is used. The answer will always be half the number you tell the person to add in the second step of the problem.

Here's another trick in which the answer will always be 1,089. Write down any three-digit number in decreasing order of digits. For example, 875 and 942 would both work. Let's use 875. Reverse the number and subtract from the original number. Therefore we have 875 - 578 = 297. Add this result to the reverse of itself. So we add 297 to 792 and we get 1,089. Try this with any three-digit number in which the numbers are from largest to smallest.

For the next trick, give someone a piece of paper and a pencil or pen. Have the person write the numbers 1 through 10 down the left hand side. Ask the person to pick two numbers between 1 and 20 and write them on lines 1 and 2. Next have the person write the sum of lines 1 and 2 and write in line 3, the sum of lines 2 and 3 and write in line 4 and so on. After the card is filled, ask the person to show the card. You should quickly be able to tell the person the sum of all 10 numbers quicker than he or she could add them on a calculator.

For example, if the person starts with the numbers 9 and 2, you can quickly tell that the sum of the numbers is 671. If the numbers chosen are 3 and 19, you can quickly say the sum is 1,837. How can this be done so fast? SImply multiply the number in line 7 by 11. If you want to check this out to show that it works, Get a card and go through the entire process. You can also astound by asking the person to divide the number in line 10 by the number in line 9. You will instantly be able to tell the person that the first three digits in that result is 1.61.

How does this work? The first result is simply based on the fact that you assign the value in line 1 the variable x and the value in line 2 the variable y. Line 3 is then x + y, line 4 is x + 2y, line 5 is 2x + 3y, line 6 is 3x + 5y, line 7 is 5x + 8y. When you get through and add all 10 lines you get 55x + 88y. That is 11 times 5x + 8y, which is line 7.

Knowing that line 10 divided by line 9 is around 1.61 is a bit more complex, involving four variables and adding fractions badly (adding numerators and denominators), showing that the value falls between two original fractions. It's not important to know how every result is arrived, but you can amaze friends by having this answer ahead of time by writing 1.61 on the back of the card.

Some of these tricks I have used over the years with my students, while the last method I obtained form the book, Secrets of Mental Math, by Arthur Benjamin and Michael Shermer.

Look for a few more mental math tricks in upcoming articles.

In previous articles, I've given you techniques needed to figure out exact answers to math problems mentally. Sometimes, we don't need an exact answer and an estimation or "best guess" is satisfactory. Suppose you are getting quotes from different banks on a personal loan. All that is needed is a close approximation to the monthly payments. Another example where an estimation is satisfactory is settling a restaurant bill with some friends, where it's not important to calculate to the exact penny. This article will explain techniques to help you master mathematical estimation.

We'll first examine addition estimation. The trick is to round the original numbers up or down. For example, 4,561 + 2,233 = 6,794. If we round to the nearest hundred, rounding up at 50 or above and rounding down below 50, we get 4,600 + 2,200 is approximately 6,800. If you always round off to the nearest hundred, the estimate will always been within 100 of the correct answer. This is within one percent of the correct answer when the answer is 10,000 or more.

We can use this technique when shopping in a supermarket. Suppose you want to approximate the total bill before the cashier rings up your order. If you round all items to the nearest 50 cents, you'd be surprised how accurate your estimation will be. Let's try it with these prices: $1.69, $2.43, $0.79, $1.57, $0.40, $4.23, $1.75, $1.35, $2.65, $0.89. The actual cost for these items is $17.75. When rounding off the prices become $1.50, $2.50, $1.00, $1.50, $0.50, $4.00, $2.00, $1.50, $2.50, $1.00. Adding these gives the estimation of $18.00, amazingly close to the total cost.


Estimation involving subtraction is done the same way. The most accurate estimation is when numbers are rounded to the nearest hundred. For example, 9,251 - 3,771 = 5,480. The estimated answer is found by taking 9,300 - 3,800 = 5,500.

For estimation involving multiplication problems, the process is similar. We still want to round numbers. If we have a multiplication problem with two, two-digit numbers, we can round each to the nearest 10. While that is quickest and easiest, it's not the most accurate. For example, take 56 times 89. We round 56 to 60 and 89 to 90 to get 60 times 90 equals 5,400. The actual answer is 4,984.


A more accurate way to estimate the problem above is to round one number up to the next 10 and the other down by the same amount the other number was round up. For example, we have 56 times 89. If we round 89 up by one to 90, we round 56 down by one to 55. Now we have 90 times 55 equals 4,950. The estimation is only 34 off from the actual answer and much easier to multiply 90 and 55 mentally than 89 and 56.

The same technique can be used when multiplying a three digit to a two digit number and two three digit numbers. Note that while the estimation technique will work, the multiplication is more difficult to perform mentally. Master the mental multiplication techniques before attempting problems above two digits.
There are other techniques for estimating answers to division problems and square roots, which will be covered in a later article. I have explained these techniques to my students over the past 14 years and hope you find some use out of these as well.

An old post on how to simplfy square roots, an important concept

In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.

The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.

For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.

Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.

When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times square root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.

Hope my method helps you when trying to figure out the square root of both positive and negative numbers.

A review of a previous topic I posted

 Simplifying square roots

In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.

The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.

For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.

Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.

When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times square root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.

Hope my method helps you when trying to figure out the square root of both positive and negative numbers.