Math 4 Final Review 3

Math 4 Final Review 3

9th - 12th Grade

19 Qs

quiz-placeholder

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Math 4 Final Review 3

Math 4 Final Review 3

Assessment

Quiz

Mathematics

9th - 12th Grade

Practice Problem

Medium

CCSS
HSN.VM.C.8, HSF.BF.B.5, HSF-BF.B.4A

+15

Standards-aligned

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19 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Given f and g in the graph,

what is f(g(2)) ?

3

8

1

4

Tags

CCSS.HSF-BF.A.1C

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Given the function g(x)=log3xg\left(x\right)=\log_3x  what is the inverse?

 g1(x)=x3g^{-1}\left(x\right)=x^3  

 g1(x)=3xg^{-1}\left(x\right)=3^x  

 g1(x)=3xg^{-1}\left(x\right)=3x  

 g1(x)=3xg^{-1}\left(x\right)=\frac{3}{x}  

Tags

CCSS.HSF-BF.B.4A

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the inverse of the function f(x)= 3(2)x5f\left(x\right)=\ 3\left(2\right)^x-5 .

 f1(x)=log2(x+53)f^{-1}\left(x\right)=\log_2\left(\frac{x+5}{3}\right)  

 f1(x)=log2(x53)f^{-1}\left(x\right)=\log_2\left(\frac{x-5}{3}\right)  

 f1(x)=log2(x+35)f^{-1}\left(x\right)=\log_2\left(\frac{x+3}{5}\right)  

 f1(x)=log2(3x5)f^{-1}\left(x\right)=\log_2\left(\frac{3x}{5}\right)  

Tags

CCSS.HSF-BF.B.4A

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How much money will in an account that has an initial principal of $5,000, and annual interest rate of 4.5% compounded monthly, and the account matures after 4 years?

$5,936

$5,983

$5,984

$5,985

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.SSE.A.1

CCSS.HSF.BF.A.1

CCSS.HSF.LE.A.1

CCSS.HSF.LE.A.2

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How much money will in an account that has an initial principal of $5,000, and annual interest rate of 4.5% compounded monthly, and the account matures after 4 years?

$5,936

$5,983

$5,984

$5,985

6.

MULTIPLE SELECT QUESTION

30 sec • 1 pt

An account has a balance of $3,000. The rate in which interest accrues in 4.1% compounded monthly. How much time in years will the money need to be invested in order to have an account balance of $3,500? Choose all that apply

 log(1+0.04112)1.16712years\frac{\log_{\left(1+\frac{0.041}{12}\right)}1.167}{12}years  

 log1.00341.16712 years\frac{\log_{1.0034}1.167}{12}\ years  

 log1.1671.003412years\frac{\log_{1.167}1.0034}{12}years  

 112log(1+0.04112)(3.53) years\frac{1}{12}\log_{\left(1+\frac{0.041}{12}\right)}\left(\frac{3.5}{3}\right)\ years  

7.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

A
B
C
D

Tags

CCSS.HSA.SSE.A.2

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