Will give brainliest to correct answer.

The Pyth. Thm. applies here:

(√x + 1)^2 + (2√x)^2 = (2√x + 1 )^2

Expanding the squares:

x + 2sqrt(x) + 1 + 4x = 4x + 4sqrt(x) + 1

Let's subtract x + 2sqrt(x) + 1 + 4x from both sides:

4x + 4sqrt(x) + 1

-(x + 2sqrt(x) + 1 + 4x)

3x + 2sqrt(x) - 4x = 0

Then 2sqrt(x) = x

Squaring both sides, 4x = x^2, or x^2 - 4x = 0. Then (x-4)x = 0, and the two possible solutions are 0 and 4.

Check these results by substitution. Does the Pyth. Thm. hold true for x=4?

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.