Answer:
A)

B)

C)

Step-by-step explanation:
We are given the function:

A)
Given that h(1) = 20, we want to find <em>k</em>.
h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

Simplify:

Anything raised to zero (except for zero) is one. Therefore:

B)
Given that h(1) = 40, we want to find 2<em>k</em> + 1.
Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

Simplify:

We can take the natural log of both sides:

By definition, ln(e) = 1. Hence:

Therefore:

C)
Given that h(1) = 10, we want to find <em>k</em> - 3.
Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

Simplfy:

Take the natural log of both sides:

Therefore:

Therefore:

Answer:
C and D
Step-by-step explanation:
So basically when u square -5 or +5, u get 25 either way
And since the square root of x is greater than or equal to 25, u can use both of them for answers.
Answer:
Formula used : (a^m)^n = a^(m×n)

- <u>d^⅝</u> is the right answer.
Answer:
$18
Step-by-step explanation:
6/10 × $30 = $18
______
Answer:
200
Step-by-step explanation:
300/15
divide cost by number bought
20
you will get this number, so for 1 ball you will pay 20
20 x 10
you will take the price for one ball and multiply it by the number of balls you want to buy
200
you will have to pay this number for 10 balls