Mathhh pl

1) which of the following function are one-to-one ? select all that apply.

a)f(x)=x-1/3x+3

b)f(x)=√5x+9

c)f(x)=7/4x^2

d)f(x)=1/2x^3

d)f(x)=3x^4+7x^3

2) find f^-1 for the function f(x)=∛x-2+8

a) f^-1(x)=(x-8)^3+2

b)f^-1(x)=(x+8)^3+2

c)f^-1(x)=∛x-8+2

d)f^-1(x)=(x-8)^3-2

5y

step-by-step explanation:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240