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Maurinko [17]
3 years ago
11

Given f(x)= (4x+1)/3 solve for f^-1(3)

Mathematics
2 answers:
nirvana33 [79]3 years ago
7 0
I hope this helps you

sineoko [7]3 years ago
6 0
F(x) = (4x + 1) /3

f⁻¹ (x ) is the inverse function of x.

Let f(x) = y,  x = f⁻¹(y)

f(x) = (4x + 1) /3

y = (4x + 1) /3                Solve for x.

3y = 4x + 1

3y - 1 = 4x          Divide both sides by 4

(3y - 1)/4  = x

x = (3y - 1)/4              Recall x =  f⁻¹(y)

f⁻¹(y)  = (3y - 1)/4   

f⁻¹(3)  = (3*3 - 1)/4

           = (9 - 1)/4 = 8 / 4 = 2

f⁻¹(3) = 2.

I hope this helps.
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Answer:

(1,1)

Step-by-step explanation:

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To find: coordinates of point H

Solution:

According to distance formula,

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So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

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Answer:

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Step-by-step explanation:

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