Rom

\(v = a t \\ v = \left(9.8 \dfrac{m}{s^2}\right)(10s) = 98 m/s\)

\(Area = length \times width \\ Area = 2\dfrac 1 4 \times 3 \dfrac 1 3 \\ Area = \dfrac 9 4 \times \dfrac {10}{3} \\ Area = \dfrac {90}{12} = \dfrac {15}{2} = 7\dfrac 1 2\)

please

first line you set a=b

thus (a-b)=0

4th line you divide by (a-b), i.e. you divide by 0.  This is undefined.

You don't have to do any of that.  They've given you the equation

s(t) = 9000 - 600t

So as in the other problem we have t as the horizontal axis, and in this case s(t) is the vertical axis.

For the vertical intercept set t=0, and evaluate s(0)

For the horizontal intercept set s(t)=0 and solve for t.

You find your GPA (in the US anyway) by setting

A=4

B=3

C=2

D=1

E or sometimes F = 0

Then take the average of your grades.

So in this case we have 2 A's, a B, and 3 C's.  6 total grades, so we sum them up and divide by 6 for the average.

4 + 4 + 3 + 2 + 2 + 2 = 17

and

17/6 = 2.83333 ...

GPAs are generally rounded to 2 digits to the right of the decimal point so your GPA is 2.83

First off you have to understand that the axes are labeled whatever they happened to be labeled so you have to be careful using terms like x-intercept and y-intercept.  In problem (b) the horizontal axis happens to be the t-axis and the vertical axis is the A axis.

Next we understand that the intercept of a given axis is then the value of the variable on the other axis is 0.

In problem (a) we find the horizontal intercept when the variable on the vertical axis, y, is 0.  We find the vertical intercept when the variable on the horizontal axis, x, is 0.

In (a) y=3x + 24

The vertical intercept is easy, just let x=0 and evaluate y.  Here we get

y=3*0 + 24 = 24

The horizontal intercept isn't quite as easy.  Here we have to set y=0 and solve for x.

0 = 3x+24

-3x = 24

x = -8

So we see the vertical intercept is at (0,24), and the horizontal intercept is at (-8, 0)

I'm going to let you work problem (b), just remember

vertical intercept: variable on horizontal axis, here t, equals 0.

horizontal intercept: variable on vertical axis, here A, equals 0.

Do you really want to live in a world where the machines are smarter than you are?

\(\mbox{point slope form is }\\ (y-y_0) = m(x-x_0) \\ \mbox{In this case }(x_0,y_0) = (1, 5) \\ \mbox{Now to find m we use }\\ m=\dfrac{y_1-y_0}{x_1-x_0} \mbox{ where }(x_1,y_1) \mbox{ is any other point known to be on the line.}\\ \mbox{In this case } (x_1,y_1) = (4,35) \mbox{ and we compute }m \mbox{ as}\\ m=\dfrac{35-5}{4-1} = \dfrac{30}{3} = 10 \\ \mbox{So our final equation is }\\ (y-5) = 10(x-1)\)

\(\mbox{If a line has slope }m_1 \mbox{ then a line perpendicular to that line will have slope } m_2 = -\dfrac 1 {m_1} \\ 7y = 9x + 1 \\ y = \dfrac 9 7 x + \dfrac 1 7\\ \mbox{This line has slope }m_1 = \dfrac 9 7 \\ -3y = x - 24 \\ y = -\dfrac 1 3 x + 8 \\ \mbox{This line has slope }m_2 = -\dfrac 1 3 \\ -\dfrac 7 9 \neq - \dfrac 1 3 \\ \mbox{So the two lines are not perpendicular.}\)

the square root of -96 doesn't exist over the real numbers.

However over the complex numbers  if you have some number 

^{\(-a, ~~~~a>0 \\ \mbox{then }\sqrt{-a} = \sqrt{-1 \times a} = \sqrt{-1}\times \sqrt{a} = \imath \sqrt{a} \\ \mbox{The symbol } \imath \mbox{ also referred to as just }i \mbox{ or in electrical engineering as }j \\ \mbox{stands for }\sqrt{-1} \mbox{ and is a Complex number.}\)}