All categories

3 years ago

What is the inverse of f(x) = sqft x+ 3

Mathematics

1 answer:

grin007 [14]3 years ago

8 0

Answer:

$f^{-1} (x)=x^{2} -3$

Step-by-step explanation:

Given the function

$f\left(x\right)=\sqrt{x+3}$

We can find the inverse function by exchanging x and y, and then solve for y.

$y=\sqrt{x+3}$

Replace x with y.

$x=\sqrt{y+3}$

square both sides

$x^2=y+3$

$y=x^2-3$

Hence,

$f^{-1} (x)=x^{2} -3$

You might be interested in

I need help with this math problem

cricket20 [7]

F(x) = 2(x + 3), now if we distribute the 2, we get 2x + 6.

we know the lbs of apple is "x", and therefore, the lbs of orange will be the constant 3.

2x will then stand for the price for "x" lbs of apples.

and 2*3 will stand for the price of "3" lbs of oranges, which is 6 bucks.

since the 2 is multiplying the lbs quantity, it must then be the price, so both cost 2 bucks each per pound.

4 0

4 years ago

Answer the following questions using what you've learned from this unit. Write your

Ket [755]

Answer:

exam

7 0

2 years ago

Round 0.8888 to 1 significant figure

Tanzania [10]

0.8888 rounded to one significant figure is 0.9

3 0

3 years ago

Use the unit circle to evaluate these expressions:

sergeinik [125]

Answer:

a) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4$

For this case we know that $sin (\pi/4) = \frac{\sqrt{2}}{2}$

So then $sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}$

b) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$19 \pi/6 - 2\pi = \frac{7\pi}{6}$

For this case we know that $cos (\pi/6) = \frac{\sqrt{3}}{2}$

And we know that $\frac{7\pi}{6}$ is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then $cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}$

c) For this case that any factor of $\pi$ the sin function is equal to 0.

So from definition of tan we have this:

$tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0$

Step-by-step explanation:

a. sin (17pi / 4 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4$

For this case we know that $sin (\pi/4) = \frac{\sqrt{2}}{2}$

So then $sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}$

b. cos (19pi / 6 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$19 \pi/6 - 2\pi = \frac{7\pi}{6}$

For this case we know that $cos (\pi/6) = \frac{\sqrt{3}}{2}$

And we know that $\frac{7\pi}{6}$ is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then $cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}$

c. tan(450pi)

For this case that any factor of $\pi$ the sin function is equal to 0.

So from definition of tan we have this:

$tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0$

4 0

3 years ago

I need it fast plss help someone help

pogonyaev

56 over 15

hope this help yoooooooooooooooooooo

5 0

3 years ago

Read 2 more answers