Veronica is ordering trophies for her school. company p charges $3.50 for each trophy and a one–time engraving fee of $25. company r charges $7.50 for each trophy and a one–time engraving fee of $17. which inequality can be used to find x, the minimum number of trophies that can be ordered so that the total charge at company p is less than the total charge at company r? a) 3.5 + 25x < 7.5 + 17x b) 3.5 + 25x > 7.5 + 17x c) 3.5x + 25 < 7.5x + 17 d) 3.5x + 25 > 7.5x + 17

The answer is C) 3.5x + 25 < 7.5x + 17

Step-by-step explanation:

the reason why is because first you have to set up the equation.

3.5 is the variable because of the keyword each and 25 isnt because that is a one time fee

this is how the company p equation should look like.

3.5x+25

same thing with company r

7.50 is the variable because of the keyword each and 17 isnt because that is a one time fee

this is how the company r equation should look like.

7.5x+17

now that you have both it says the total charge of company p when its less than p

now it should look like this

C) 3.5x + 25 < 7.5x + 17

and thats how u get the answer

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.