hectictar

Agree completely....thank you for posting this ♥

$\phantom{=\qquad}(\ 3\sqrt6\ \text{cis}(\frac{\pi}{8})\ )(\ 2\sqrt5\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \text{cis}(\frac{\pi}{8})\ )(\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})+i\sin(\frac{\pi}{8})\ )(\ \cos(\frac{7\pi}{6})+i\sin(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) } \ +\ i {\color{purple}(\ \cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ \sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )} \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8}+\frac{7\pi}{6}) } \ +\ i\ {\color{purple} \sin(\frac{\pi}{8}+\frac{7\pi}{6}) } \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \cos(\frac{31\pi}{24}) } \ +\ i\ { \sin(\frac{31\pi}{24}) } \ )\\~\\ {=\qquad}6\sqrt{30}\ \text{cis}(\ \frac{31\pi}{24}\ )$

x  =  t - 2

y  =  2t² + 4

Let's solve the first equation for  t  and substitute that in for  t  into the second equation.

x  =  t - 2     Add  2  to both sides of this equation.

x + 2  =  t

Since  t  =  x + 2  we can substitute   x + 2  in for  t  into the second equation.

y  =  2t² + 4

y  =  2(x + 2)² + 4

I'll let you expand the right side of that equation if necessary.

To find the smallest possible value of  x,  plug in the smallest possible value of  t  into the equation   x = t - 2

And to find the biggest possible value of  x, plug in the biggest possible value of  t  into the equation   x = t - 2

When   t  =  -3,     x   =   -3 - 2   =   -5

When   t  =   1,     x   =    1 - 2   =   -1

So  x  is in the interval  [-5, -1]

Since the point  (3, 6)  is on the graph of  y = g(x)  then  g(3)  =  6

And...

h(3)  =  ( g(3) )²   =   (  6  )²   =   36

Since  h(3) = 36  then the point  (3, 36)  is on the graph of  y = h(x)

3 + 36  =  39

area of triangle   =   $\frac12$  *  base  *  height

We know the base is  4 inches.  To find the height we can use the Pythagorean theorem.

4²  +  h²   =   √[ 97 ]²

16  +  h²   =   97

h²   =   81

h   =   9

The height is  9  inches.

base   =   4 inches   =   4/12  feet   =   1/3  feet

height   =   9 inches   =   9/12 feet   =   3/4 feet

area of triangle   =   $\frac12\cdot\frac13\cdot\frac34\ $   sq feet   =   $\frac18$  sq feet

〈 -7, -3 〉  +  〈 -2, 10 〉   =   〈 -7 + -2, -3 + 10 〉   =   〈 -9, 7 〉

To get from the starting point to the terminal point, go left 2 units and down 4 units.

So the vector is  <-2, -4>

%   means the same thing as  /100

3.333333   =   3.333333 * 100 / 100   =   333.3333 / 100   =   333.3333 %

3.333333  expressed as a percent is   333.3333%

Also...

For these,

You can replace  U  with  "or"

And replace  ∩  with  "and"

For example...

P(owns a PS4  U  does not own a cellphone) is the probability that a randomly selected person in the group owns a PS4 or does not own a cellphone.