Question 22 of 40 The table shows values for a quadratic function.
X
y
0
0
1
2
2
8
3
18
4
32
5
50
6
72
What is the average rate of change for this function for the interval from x= 3
to x = 5?
O A. 16
B. 25
Ο Ο Ο Ο
O C. 10
A
O D. 32
PREVIOUS

it is true

step-by-step explanation:

all the combinations add up i used   calculator

c.   x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

.

to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!