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Algebra 8 Online
OpenStudy (anonymous):

The sides of a square are 2 to the power of 4÷9 inches long. what is the area of the square?

Vocaloid (vocaloid):

side = 2^(4/9) area = side^2 = ?

OpenStudy (anonymous):

Wait what

OpenStudy (anonymous):

How can I solve this??? @vocaloid

Vocaloid (vocaloid):

it gives you the length of one side, so to find the area, you just square the length (side)*(side) = 2^(4/9) * 2^(4/9) = ?

OpenStudy (anonymous):

Do I times it 4 times or just twice @vocaloid

Vocaloid (vocaloid):

twice..

OpenStudy (anonymous):

@vocaloid

OpenStudy (anonymous):

I got 4 to the power of 16/81

Vocaloid (vocaloid):

not quite, we have to use the exponent rule here keep the base and add the exponents together 2^(4/9) * 2^(4/9) = 2^(4/9 + 4/9) = ?

OpenStudy (anonymous):

Oh ok

OpenStudy (anonymous):

Now I got 2 to the power of 8/9

OpenStudy (anonymous):

Is that right? @vocaloid

Vocaloid (vocaloid):

yes, sorry for the late reply

OpenStudy (anonymous):

Do you you think you can help me with another question? @vocaloid

Vocaloid (vocaloid):

sure

OpenStudy (anonymous):

Given the function f(x) = 5^x , section A is from = x = 0 to x = 1 and the section B is from x = 2 to x = 3. Part A: Find The average rate of change of each section Part B: How many times greater is the average rate of change of section B then section A? Explain why one rate of change is greater than the other. @vocaloid

Vocaloid (vocaloid):

part A) we want two values 1. the rate of change between x = 0 and x = 1 and 2. the rate of change between x = 2 and x = 3 so, let's just tackle the first part for now the rate of change between x = 0 and x = 1 can be found using the formula [f(1)-f(0)]/(1-0)

OpenStudy (anonymous):

I got one

Vocaloid (vocaloid):

not quite, let's take it bit by bit f(1) = ?

OpenStudy (anonymous):

Is it one

Vocaloid (vocaloid):

not quite f(x) = 5^x f(1) = ?

OpenStudy (anonymous):

5 to the power of one

Vocaloid (vocaloid):

right now f(0) = ?

OpenStudy (anonymous):

5 to the power of zero

Vocaloid (vocaloid):

right, so [f(1)-f(0)]/(1-0) = ?

OpenStudy (anonymous):

Is it 1/1

Vocaloid (vocaloid):

no... remember, we just calculated f(1) and f(0) f(1) - f(0) = ?

OpenStudy (anonymous):

5^1 - 5^0

Vocaloid (vocaloid):

right, so [f(1)-f(0)]/(1-0) = ?

OpenStudy (anonymous):

Do I have to solve the exponent

Vocaloid (vocaloid):

yes

OpenStudy (anonymous):

5-1/ 1-0 = 4/1

Vocaloid (vocaloid):

right, and 4/1 = ?

OpenStudy (anonymous):

4

Vocaloid (vocaloid):

right, now let's calculate the rate of change from x = 2 to x = 3 [f(3)-f(2)]/(3-2) = ?

OpenStudy (anonymous):

5^3 - 5^2/ 3-2

Vocaloid (vocaloid):

right, keep going...

OpenStudy (anonymous):

125 - 25/ 3-2

Vocaloid (vocaloid):

right, keep going...

OpenStudy (anonymous):

100/1= 100

Vocaloid (vocaloid):

right

Vocaloid (vocaloid):

now we're finished w/ part A part B wants us to find (rate of change from x = 2 to x = 3)/(rate of change from x = 1 to x = 2) using what we calculated earlier... 100/4 = ?

OpenStudy (anonymous):

25

Vocaloid (vocaloid):

right so section B is 25 times as great (24 times greater) as section A

OpenStudy (anonymous):

So that's the answer

Vocaloid (vocaloid):

as for "explain why" I would say something along the lines of "f(x) is an increasing exponential function

Vocaloid (vocaloid):

yeah

OpenStudy (anonymous):

Thank you

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