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That is the 'standard form' equation for a line.

"what is my score out of 100% if 18 was max and i got 10."

score as a percentage:     100*10/18 → 55.6%

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2.5+2.5=5

"Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. x = 7y2, x = 7; about x = 7"

"When a 49 pound weight is attached to a spring, the spring stretches 7 inches. Assume that the length of the spring y varies directly withthe weight attached x. Write and solve a direct variation equation to find the length of the spring when a 63 pound weight is attached"

Direct variation means:   y = k*x   where k is a constant

Find k by using the given values of y = 7 inches and x = 49 pounds:     7 = k*49   so k = 1/7 inches per pound

Now, when x = 63 pounds we have:   y = (1/7)*63 inches or y = 9 inches

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3/(x-2)-1/(x+2)=5

\(\begin{array}{|rcll|} \hline 5 &=& \frac{3}{x-2}-\frac{1}{x+2} \quad & | \quad \cdot (x+2)(x-2) \\ 5\cdot(x+2)(x-2) &=& \frac{ 3(x+2)(x-2)}{x-2}-\frac{(x+2)(x-2) }{x+2} \\ 5\cdot (x+2)(x-2) &=& 3(x+2) - (x-2) \\ 5\cdot (x+2)(x-2)) &=& 3x+ 6 - x+2 \\ 5\cdot (x+2)(x-2) &=& 2x+8 \\ 5\cdot (x^2-4) &=& 2x+8 \\ 5x^2-20 &=& 2x+8 \quad & | \quad -2x-8 \\ 5x^2-20 -2x-8 &=& 0 \\ 5x^2-2x -28 &=& 0 \\ \\ \begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{-b\pm\sqrt{b^2-4ac} }{2a} \\ \hline \end{array} \\ ax^2+bx+c &=& 0 \quad & | \quad a = 5 \quad b = -2 \quad c=-28 \\ x &=& \frac{ 2\pm\sqrt{4-4\cdot 5 (-28)} }{2\cdot 5} \\ x &=& \frac{2\pm\sqrt{4+20\cdot28)} }{10} \\ x &=& \frac{2\pm\sqrt{564} }{10} \\ x &=& \frac{2\pm 23.7486841741 }{10} \\\\ x_1 & = & \frac{2+ 23.7486841741 }{10} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2.57486841741 } \\\\ x_2 &=& \frac{2- 23.7486841741 }{10} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-2.17486841741} \\ \hline \end{array} \)

Thou shalt not divide by zero.

The operation of dividing by zero is not defined, you are not allowed to do it.

0/0 is meaningless.

2. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate sin 2 θ.

3. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate cos 2 θ.

\(\begin{array}{|rcll|} \hline r^2 &=& (4-r)^2+3^2 \\ r^2 &=& 4^2-8r+r^2+3^2 \\ 8r &=& 4^2+3^2 \\ 8r &=& 5^2 \\ r &=& \frac{5^2}{8} \\ r &=& \frac{25}{8} \\ \hline \end{array}\)

2. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate sin 2 θ.

\(\begin{array}{|rcll|} \hline \sin(2\theta) &=& \frac{3}{r} \\ \sin(2\theta) &=& \frac{3}{\frac{25}{8}} \\ \sin(2\theta) &=& \frac{24}{25} \\ \hline \end{array}\)

3. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate cos 2 θ.

\(\begin{array}{|rcll|} \hline \cos(2\theta) &=& \frac{4-r}{r} \\ \cos(2\theta) &=& \frac{4}{r}-1 \\ \cos(2\theta) &=& \frac{4\cdot 8}{25}-1 \\ \cos(2\theta) &=& \frac{32}{25}-1 \\ \cos(2\theta) &=& \frac{7}{25} \\ \hline \end{array}\)

how do I find the radius and the arc length?

From a surface A you calculate

\(r= \frac{A}{2\pi}\)

\(c=2\pi r\)

!

It is 12

           C

              3   that is (12x11x10)/(3x2x1)

because  

number of choices for first book =12  which leaves

number of choices for 2nd book=11    which leaves

number of choices for 3rd book =10.                            With each of your first choices of 12 you then have another 11 choices to make  which is 12x11,and with each of those you have another 10,giving you 12x11x10 choices.  BUT  !!!!   -- you aren't finished yet,because within that number of 12x11x10 choices,you will have picked the same bunch  of books more than once.  For example if you picked   books (call them books 1,2 and 3) you will also have picked 132, 213,231,312,321,which are all the same bunch of books. That is you will have picked the same  bunch (called a combination) of 3 books six times.  So you need to divide the number of choices by the number of ways (called permutations) you can arrange each choice. Which is six.

A combination is a collection of objects (like your bunch of 3 books),a permutation is the number of ways you can arrange those objects.

How to find the inverse of a cubic function?

\(\begin{array}{|lrcll|} \hline 1. & y &=& 2(x-3)^3-16 \\\\ \text{Inverse:} & x &=& 2(y-3)^3-16 \quad & | \quad +16 \\ & x+16 &=& 2(y-3)^3 \quad & | \quad :2 \\ & \frac{x +16}{2}&=& (y-3)^3 \\ & \frac{x}{2} + 8 &=& (y-3)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{\frac{x}{2} + 8} &=& y-3 \quad & | \quad +3 \\ & \sqrt[3]{\frac{x}{2} + 8}+3 &=& y \\ & y &=& \sqrt[3]{\frac{x}{2} + 8}+3 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & f(x) &=& \sqrt[3]{x + 1}+5 \\\\ \text{Inverse:} & x &=& \sqrt[3]{y + 1}+5 \quad & | \quad -5 \\ & x-5 &=& \sqrt[3]{y + 1} \quad & | \quad \text{cube both sides} \\ & (x-5)^3 &=& y + 1 \quad & | \quad -1 \\ & (x-5)^3-1 &=& y \\ & y &=& (x-5)^3-1 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline 3. & y &=& \frac12(x-1)^3+3 \\\\ \text{Inverse:} & x &=& \frac12(y-1)^3+3 \quad & | \quad -3 \\ & x-3 &=& \frac12(y-1)^3 \quad & | \quad \cdot 2 \\ & 2(x-3) &=& (y-1)^3 \\ & 2(x-3) &=& (y-1)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{2(x-3)} &=& y-1 \quad & | \quad +1 \\ & \sqrt[3]{2(x-3)}+1 &=& y \\ & y &=& \sqrt[3]{2(x-3)}+1 \\ \hline \end{array}\)

1/(x+3)-1/(x-3)=3

\(\begin{array}{|rcll|} \hline 3 &=& \frac{1}{x+3}-\frac{1}{x-3} \quad & | \quad \cdot (x+3)(x-3) \\ 3\cdot (x+3)(x-3) &=& \frac{ (x+3)(x-3)}{x+3}-\frac{ (x+3)(x-3)}{x-3} \\ 3\cdot (x+3)(x-3) &=& (x-3) - (x+3) \\ 3\cdot (x+3)(x-3) &=& x- 3 - x-3 \\ 3\cdot (x+3)(x-3) &=& -6 \quad & | \quad : 3 \\ (x+3)(x-3) &=& -2 \\ x^2-9 &=& -2 \quad & | \quad +9 \\ x^2 &=& 7 \quad & | \quad \text{square root both sides} \\ \mathbf{x} & \mathbf{=} & \mathbf{\pm\sqrt{ 7 } }\\ \hline \end{array}\)