Which function can be used to represent the graphed geometric sequence?

f(x) = 80()
f(x) = 320(
f(x) = 80(4)x – 1
f(x) = 320(4)x – 1

f(x) = 80(1/4 x^x-1) ⇒ first answer

Step-by-step explanation:

* Lets explain the geometric sequence

- In the geometric progression there is a constant ratio between

 each two consecutive numbers

Ex:

5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric Progression:

- U1 = a  ,  U2  = ar  ,  U3  = ar2  ,  U4 = ar3  ,  U5 = ar4

∴ Un = a(r)^n-1, where a is the first term , r is the constant ratio

  between each two consecutive terms  and n is the position of

  the number in the sequence

* In the problem we have 4 answers we will search which one can

 be put in the form a(r)^n-1

- The first answer is 80(1/4 r^x-1)

∵ We can multiply 80 by 1/4 ⇒ 80 × 1/4 = 20

∴ It will be 20(x^x-1) ⇒ the form of the geometric sequence

* The answer is the first answer

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

the answer is b