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2 answers:

astra-53 [7]3 years ago

7 0

I think the answer is c and the reason why is

babunello [35]3 years ago

6 0

whatever that person said

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What is the inverse of f(x)=(x+6)2 for x≥–6 where function g is the inverse of function f? g(x)=x√−6, x≥0 g(x)=x−6−−−−√, x≥6 g(x

rosijanka [135]

Answer:

$g(x)=\sqrt{x}-6$ , x>=0

Step-by-step explanation:

$f(x)=(x+6)^2$

To find the inverse function , replace f(x) by y

$y=(x+6)^2$

Replace x with y and y with x

$x=(y+6)^2$

Solve the equation for y

take square root on both sides

$+\sqrt{x} =y+6$

Now subtract 6 from both sides

$+\sqrt{x}-6=y$

replace y with g(x)

$g(x)=\sqrt{x}-6$

7 0

3 years ago

Please solve will give brainliest

Over [174]

Answer: 12

Step-by-step explanation: Since 11 is ruled out, answer is 12 people

4 0

3 years ago

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.