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Free_Kalibri [48]
1 year ago
11

Find the inverse of each function. Is the inverse a function? f(x)=2/3 x-3

Mathematics
1 answer:
kykrilka [37]1 year ago
7 0

The inverse for the function f (x) = 2 / 3 x - 3 will be f ⁻¹ (x) = 3 / 2 x + 9 / 2.

We are given the function:

f (x) = 2 / 3 x - 3

we need to find the inverse of the function.

First we will write the function in the form of x.

y = 2 / 3 x - 3

y + 3 = 2 / 3 x

2 x = 3 y + 9

x = 3 / 2 y + 9 / 2

So, the inverse function will be:

f ⁻¹ (x) = 3 / 2 x + 9 / 2.

Therefore, the inverse for the function f (x) = 2 / 3 x - 3 will be f ⁻¹ (x) = 3 / 2 x + 9 / 2.

Learn more about inverse function here- brainly.com/question/18226349 "

#SPJ4

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What is the exterior angle?
olganol [36]

Answer:

angle YUT = 120

Step-by-step explanation:

in Triangle SUT

angle 70 + angle 50 + angle x = 180

 70 + 50 + x = 180

 120 + x = 180

 x = 180 - 120

 x = 60

angle SUT = 60



angle SUT + angle YUT = 180 ( straight line)

60 + x = 180

 x = 120


Therefore angle YUT ( exterior angle ) = 120

7 0
3 years ago
Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1
erik [133]

Answer:

\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

Step-by-step explanation:

Given that:

y =  x^{2/3} at y = 0 , x = 1

Then:

Area = \int^{1}_{0} x^{2/3} \ dx

Area = \begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0

Area = \dfrac{3}{5}

Then:

\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx

\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0

\overline x = \dfrac{5}{3} \times \dfrac{3}{8}

\overline x = \dfrac{5}{8}

Similarly;

\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2}  (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14}  (x^{7/3} ) \end {bmatrix} ^1_0

\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )

\overline y = \dfrac{5}{14}

Thus; \mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

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