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3 years ago

5

1) Find the inverse of the function below. Use g-1(x) to indicate the inverse 5 points

1 answer:

omeli [17]3 years ago

4 0

Answer:

g-1(x)=SQRT{(x-4)/2} + 4

Step-by-step explanation:

I assume that the 2 is the exponent so ^ represents exponent

We know your function is g(x)=2(x-3)^2+4

For inverse functions we swap x and y values, note g(x) is like the y value

Isolate for y

y=2(x-3)^2+4

x=2(y-3)^2+4

(x-4)/2 = (y-3)^2

SQRT{(x-4)/2} + 4 = y

Therefore g-1(x)=SQRT{(x-4)/2} + 4

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Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

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Answer:

$\sin L = 0.60$

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Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

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Answer:

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