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Ok nvm CPhill got it. Sorry for any confusion I might have caused.

Thanks for the answer!!! Is there any direction you can point me to learn more about finding the answer when they ask for the value of f(x)=-3 i would really like to learn that

Try to find a circle surface.

1.  Here's the graph : https://www.desmos.com/calculator/sr1qh8cxfs

f(x) = - 3

We want to know  what values of x produce a y value of - 3

The graph shows three  points where this is true  (-4.5, - 3)  ( -2, - 3)  and (12, - 3)

2.   The parent graph is   f(x) = l x l      ( absolute value graph )

It wil have a "V" shape

f(x)  = -3 l x + 1 l - 2

The "- " reflects the graph over the x axis

The "3"  vertically stretches the graph by a factor of 3

"x + 1"  indicates that the graph is shifted to the left by 1 unit

The " -2 "  shifts the graph downward by 2 units

I can't help with the second part, for I am not to good at functions. In the first part, you have three equations saying, if x is between these two numbers, then this is the equation. So it tells us that x = -3, so we know that it's the second equation. So just use a graphing calculator to graph the function f(-3) = (x+2)² -3

If I'm reading this right, you gotta just solve each equation indiviudualy. And when you have negative exponents, you take the inverse of the number, then the exponent turns positve. So for example, 2^-1 = 1/2

Note, GM.....

a^-b  =  1 / a^b

So

2^-3 =  1/2³  = 1/8

Can you figure out the next three based on this???......post your answers and I'll check them

[ Also....    if a is not 0.....   a⁰  =  1 ]

The last  one is 2³  =  2 * 2²   =  2 * 4 = 8

I'm not as well-versed in this as EP....but....I believe that we are using something known as Charles's Law that says :

V₁ T₂  = V₂ T₁        so

620 * 325 = V₂ * 293      divide both sides by   293

(620 * 325) / 293 = V₂   ≈ 687.7 ml

r  =  sqrt ( 2^2 + (√21)^2 )  =  sqrt ( 4 + 21)  = sqrt (25) = 5

 θ = arctan ( √21 / 2) ≈  66.4°  

5cos(66.4°) + 5i sin (66.4°)   =  5 [ cos (66.4°  + i sin (66.4° ) ]

z =  8( cos20 + isin 20)

w = 4 (cos60 + isin 60)

z * w =

32   (  [ cos20cos60 + [sin 20cos60] i  + [ sin 60cos20 ] i  +  [ sin 20 sin 60 ] i^2     )  =

32  ( [ cos20cos60 ] + [sin 20cos60] i  + [ sin 60cos20 ] i  -  [ sin 20 sin 60 ]    )  =

32    (   [  cos20cos60 - sin 20sin 60]  +  [ sin20cos 60 + sin60cos20] i    )   =

32 (   cos ( 20 + 60) + sin (20 + 60)  i  )  =

32 ( cos (80)     +  i sin ( 80)  )

y = -10

r sin  θ =  -10

r =  -10 / sin  θ

r =  -10 csc  θ

Great response! I have somewhat of an understanding of how you got the answer 1. I have also realized from this question that the materials necessary to solve it have not been taught at the grade 10 level, and am a little mad at my math teacher. I really was stuck on this one, great explanation once again!

Correct-a-mundo !    Just hoping to avoid confusion for those learnig the system....   

Note that temperatures in Kelvin do not use degrees. The unit measure is simply Kelvin(s).

Eh.....L'Hopitals  is just as good....and probably more "compact"

Like I always say.....there may be many roads that lead to the same destination !!!!

cos 315°   =   45° reference angle in the 4th quadrant =  √2 /2

And sin 315°    is the negative of this

So

z = (3) cos 315°  + ( 3i ) sin 315°  =

[   (3/2)√2  ,  - (3/2) √2 i  ]

Here is the graph

I actually like your way better.1...I just happened across l'Hopital and thought I would use it....    Thanx  !  ~EP

Thanks EP....here's another way not using L'Hopital's

lim                   sin (9x)

x → 0             ______              multiply top/bottom by 9

                          x

lim x → 0            9 sin (9x)

                       _________                let 9x =     θ

                              9x

lim x → 0          9 sin ( θ)

                        ________

                               θ

9   *       lim x → 0       sin (  θ )

                                  _______      =

                                        θ

9  *   1      =

9

L'Hopitals Rule

Lim x ----.>0  (cos(9x) *9)/1  = lim x----> 0  9 cos(0) = 9

Ideal gas law  

PV = nRT

P/nR = T/V     in your question's situation   P is consatnt   n is constant and R is the ideal gas constant....so the entire left side of this equation is a constant....let's call it  ' k '

so

k = T1/V1  = T2/V2       when dealing with gas laws   T MUST BE IN Kelvin degrees !!!

20 C = 273.15 +20  K   = 293.15K

soooo....

293.15 k / 2L  =  x/ 1L        x is the temp (in Kelvin we are looking for)  

          x = 293.15/2 = 146.58 k          146.58-273.15 =  - 126.58 C

You can just calculate it out to find the coordinates:

3 cos 315 = 2.12

3 i sin 315 = -2.12 i

plot    2.12 - 2.12 i     on the complex plane....

Is that supposed to be    150 ?

I think      

Given   

-3 +3i          it should be       4.24 cos 135   + i 4.24 sin 135      DO you have some numbers incorrect?

Oh now I see....your question header and your question text are different

Given   -3 sqrt(3)  + 3 i                   |z| = r =sqrt [ (-3sqrt3)^2   + 3^2 ]  = 6

then   6 cos 150 = -3 sqrt3      and    6i sin 150 = 3i      SO the points are the same on the graph.....

The complex plane looks like the usual cartesian coordinate plane....but the   x axis is for the REAL part of your complex number and the imaginary part is plotted on the ' i ' axis  (y-axis)....your graphed point would look like this (note the labels on the axes):

"The highest point in a particular section of a graph"

Here is a graph of a function with a lot of relative maximums: