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2 years ago

Find the inverse of the function y = 3x + 5 Find the inverse of the function y=3x+5

Mathematics

1 answer:

Viktor [21]2 years ago

4 0

Answer:

First Inverse function= f^-1(x)=x/3-5/3

Second Inverse function=f^-1(x)=x/3-5/3

Step-by-step explanation:

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Can someone pls help for brainlest

uranmaximum [27]

Answer:

a = 6

Step-by-step explanation:

Area = 36 $in^{2}$

Only one number will make this missing length true.

Obviously it's a square so one number x one number will equal to 36 $in^{2}$

6 x 6 = 36 $in^{2}$

7 0

3 years ago

Read 2 more answers

How do you find and use the greatest common factor of two whole numbers

Tasya [4]

For example you have the 52, and you have the 24, you look for a number that can go into both that is the largest possible for my example it would be 4 since no number greater can go into both, while 2 would be a option it would not be the greatest.

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3 years ago

Multiply 25 x 47 x 3

dmitriy555 [2]

3525 is your answer please consider brainliest<3

6 0

3 years ago

Easy question.. giving brainliest<br> x/16 = 10/36<br> x=_

irina [24]

Answer:

x=40/9

Step-by-step explanation:

$\frac{x}{16} = \frac{10}{36} |first \: you \: cross \: multiply| \\ x(36) = 16(10) |open \: bracket | \\ 36x = 160 |divide \: both \: side \:by \: 36 | \\ \frac{36x}{36} = \frac{160}{36} \\ x = \frac{40}{9}$

6 0

3 years ago

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

3 years ago

Find the inverse of the function y = 3x + 5 Find the inverse of the function y=3x+5​

Find the inverse of the function y = 3x + 5 Find the inverse of the function y=3x+5