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3 years ago

Inverse Functions Question: Find the inverse for f(x)= 6x+15

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

Answer: Finding the Inverse of a Function

First, replace f(x) with y . ...

Replace every x with a y and replace every y with an x .

Solve the equation from Step 2 for y . ...

Replace y with f−1(x) f − 1 ( x ) . ...

Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.

Step-by-step explanation:

nikitadnepr [17]3 years ago

7 0

Answer:

attached pic

Step-by-step explanation:

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There were 12 people on jury there were 4 more men than women how many men were there

AlekseyPX

Answer:

8 men.

Step-by-step explanation:

12÷3=4.

4 women, 8 men.

8-4=4.

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4 years ago

PLEASE HELP!! Convert each linear equation to standard form. y-4= -2(x+3)

hjlf

So first, this is point slope form. The first step is to distribute.)
$y-4=-2x-6$
Then add 4 to both sides.
$y=-2x-2$
Then add 2x and it becomes
$2x+y=-2$

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4 years ago

Read 2 more answers

What value of t is the solution of the above equation?

mel-nik [20]

Answer:

15/4, or 3 3/4 is your answer.

Step-by-step explanation:

Isolate the variable, t. Note the equal sign, what you do to one side, you do to the other. Multiply 3/2 to both sides.

(3/2)(2/3)t = (5/2)(3/2)

t = (5/2)(3/2)

t = 15/4

15/4, or 3 3/4 is your answer.

7 0

4 years ago

Economic Leading Indicators are based on historical results. A.True B False

deff fn [24]

I think it’s A I’m not 100 percent sure tho sorry if I’m wrong

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3 years ago

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

8 0

3 years ago