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Wait, how did you get the first summation? Is there a formula or something?

See the last one, here :

https://web2.0calc.com/questions/study-guide-help-asap-for-act

Perimeter  =  42 * 0.5 cm  =   21 cm

Area = 34 * (0.5)^2  =  34 * .25  =   34  * (1/4)  = 34/4  = 17/2  = 9.5 cm^2

1. 

a    radii    = AC  and AD           tangents = EC and ED

b      AED forms a right triangle with hypotenuse AC = 17  and leg AC = 8

        Using the Pythagoren Theorem,  DE  =  sqrt [ AC^2 - AC^2 ] = sqrt [ 289 - 64 ] =

        sqrt [ 225]  =  15

2.  B is translated up 1 unit and to the left 9 units

     The scale factor  is 2

      Don't know what "conclusion"  they want

3.   Angles O and Q are supplementary...so

        (x + 17) + ( 6x - 5) =  180

         7x + 12   =  180      subtract 12 from both sides

           7x  =  168    divide both side by 7

            x = 24

P and R are also supplementary...so

y + (2x + 19)  =  180

y +  (2*24 + 19) = 180

y + 67  = 180      subtract 67  from both sides

y = 113

Angle O  =   x + 17 =  24 + 17  = 41 degrees

Angle Q =  6(24) - 5  =  139 degrees

Angle P = y =  113 degrees

Angle R =  2(24) + 19 =  67 degrees

Arc RP  =  2 * measure of angle O  =  2 * 41   =   82 degrees

First  one

Rate of change from 5 to 9  =  f(9) - f(5)            14  - (-2)            16 

                                               _________  =     ________ =      ___     =   4

                                                   9  -   5                   4                    4

Second one

( f/g)(x)  means   f(x)            4x^2  +  6x                      2x ( 2x + 3)             2x

                         ____  =       ______________  =     _____________  =  _____

                           g(x)            2x^2 + 13x + 15           (2x + 3) (2x +5)        2x + 5

Last one

x                 x^2 - 6x + 8               x - 2

0                           8                        -2

1                           3                        -1

2                           0                         0

3                           -1                        1                      

4                            0                         2

5                            3                        3            

The solutions occur at x = 2  and x = 5

$x^k,~k<0 \sim \dfrac{1}{x^k},~k > 0\\ \text{It should be pretty clear the range of this is }(0,1], \text{ over a domain of }[1,\infty)$

sorry... can you please explain? i dont quite understand. i dont think you are wrong?

We don't know what $x$equals...

1)

Let  f(x)=x^2+3x+2 .

What is the average rate of change of f(x) from 3 to 6?

f(3)  =  (3)^2 + 3(3) + 2  =   20

f(6)  = (6)^2 + 3(6) + 2  = 56

Avg rate of change from 3 to 6  =  [ f(6) - f(3) ]             56 - 20          36

                                                     ___________  =     _______  =    ___   =  12

                                                          6  -   3                     3                 3

2)  Avg rate of change  from - 2 to 2  =  [ 9 - 25 ]           -16

                                                               _______ =       ____  =  - 4

                                                                 2 - (-2)               4

Not sure what you want....but

6a )   Here is a graph with one solution : https://www.desmos.com/calculator/4zppt4c1pe

6b )  Here is a graph with two solutions : https://www.desmos.com/calculator/tpgevlpwth

6c)   No solutions : https://www.desmos.com/calculator/rdyrpnzvn8

6d )  We can't have both a linear graph and a quadratic graph with infinite solutions...

ok :)

If you know that the speed you travel in a car is

miles per hour    (if you are from the Uited States)    or

kilometers per hour  from most other places then you know that

Speed = distance per time

and 'per' means divide

so

speed = distance/time

Thanks Rom.

Thanks

1) is the domain of function A is the set of real numbers greater than 0

        the domain of function B is the set of  real numbers greater than or equal to 1

2) f(-2) = g(-2)

3) the distance plane A traveled in 1h is greater than the distance plane B traveled in 1h

4) Function A has a vertical asympotote at x=3

    Function A has a vertical asympotote at x= 4

5) both functions have one negative and one positive x-intercept

$\left .\sum \limits_{k=1}^n ~k^2 = \dfrac 1 6 n (1+n)(1+2n) \right |_{n=2016} = \\ \dfrac 1 6 (2016)(2017)(4033)$

$\dfrac 1 6 (2016)(2017)(4033) \pmod{17} = \\ (336 \pmod {17})(2017 \pmod{17})(4033 \pmod{17}) \pmod{17} = \\ 13 \cdot 11 \cdot 4 \pmod{17} = \\ 572 \pmod{17} = 11$

Hello Melody: Mathematica 11 Home Edition. I actually expressed it like this: ∑[n^2, n, 1, 2016] mod 17 = 11.

But, I also re-checked it using C++ programming language.

no it did the time was messed up that i did on my own work i kept messing up on the time i thought it was right until my sis pointed out that was using the wrong formula

Thanks for responding Cerenetie.

I am glad that you worked it out :) 

I hope my answer made you think a bit even though you went to your sister for more help :)

What calc did you put that into?

so after getting help from my sister i realized i wasn't focusing on the time that the two planes took to travel the distances. plane A was faster at traveling the distance than plane B. the reason plane A was faster is because Plane A has a speed of 290 miles/hour while plane B has a speed of 270  miles per hour. for plane B i had to use the formula d=rt,  im not sure how to type my solution on here. this makes plane A faster than Plane B so the answer would be The distance plane A traveled in 1h is greater than the distance plane B traveled in 1 hour

sumfor(n, 1, 2016, (n^2) = 2733212496 mod 17 = 11

Yes that  is correct.

$\dfrac{\frac 7 8 \zeta(3)}{\zeta(3)} = \dfrac 7 8$

I think that you have already been given the answers to many of these. 

So it is time for you to start learning how to do them.

I will give you some examples that will help.

$5^{1/2}=\sqrt5\\ 7^{-3}=\frac{1}{7^{+3}}\\ \\~\\ 8^{3/2}=(8^3)^{1/2}=(8^{1/2})^3$

Now think about it and you tell me what  these are equal to.

$1) \;\; 2^{-1/2}\\ 2) \;\; 2^{-3/2}\\ and \\ 3) \;\;b^{1/2}$

Please no one answer over me