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2 years ago

What is the inverse of the function f(x) = 4x - 12

Mathematics

2 answers:

tensa zangetsu [6.8K]2 years ago

4 0

Answer:

y=x+12/4

Step-by-step explanation:

y=4x-12

x=4y-12 (then make y the subject)

x+12=4y(divide both sides by 4)

y=x+12/4

Brrunno [24]2 years ago

4 0

Y=x+12/4.
There you go bud

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Lerok [7]

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3) yes
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3 years ago

-57 + 57 I really need the answer

3241004551 [841]

Answer:

The answer is 0

Step-by-step explanation:

-57 cancels out 57 making it 0.

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6 months ago

Read 2 more answers

Help me answer this question

borishaifa [10]

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2 years ago

Please help logarithms!

nlexa [21]

Given:

$\log_34\approx 1.262$

$\log_37\approx 1.771$

To find:

The value of $\log_3\left(\dfrac{4}{49}\right)$ .

Solution:

We have,

$\log_34\approx 1.262$

$\log_37\approx 1.771$

Using properties of log, we get

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349$ $\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37$ $[\log x^n=n\log x]$

Substitute $\log_34\approx 1.262$ and $\log_37\approx 1.771$ .

$\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)$

$\log_3\left(\dfrac{4}{49}\right)=1.262-3.542$

$\log_3\left(\dfrac{4}{49}\right)=-2.28$

Therefore, the value of $\log_3\left(\dfrac{4}{49}\right)$ is $-2.28$ .

5 0

2 years ago

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an expone

ad-work [718]

Answer:

Part 1) The exponential function is equal to $y=1,350(0.95)^{x}$

Part 2) The population in 2010 was $992\ fish$

Step-by-step explanation:

Part 1) Write an exponential decay function that models this situation

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

y ----> the fish population of Lake Collins since 2004

x ----> the time in years

a is the initial value

b is the base

we have

$a=1,350\ fish$

$b=(100\%-5\%)=95\%=0.95$

substitute

$y=1,350(0.95)^{x}$ ----> exponential function that represent this scenario

Part 2) Find the population in 2010

we have

$y=1,350(0.95)^{x}$

For $x=(2010-2004)=6\ years$

substitute

$y=1,350(0.95)^{6}=992\ fish$

7 0

2 years ago