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3 years ago

Which function is the inverse of f(x) = 2x + 3 ?

Mathematics

1 answer:

siniylev [52]3 years ago

6 0

I think it might be f^-1 (x) = x/2 - 3/2
I hope this helps!

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PLEASE HELP ME ! IMAGE IS BELOW IF YOU GET IT RIGHT I'LL MARK BRAINLIEST

jarptica [38.1K]

Answer:

x = 17

y = 8

QPR = 35º

QRP = 60º

Step-by-step explanation:

2x + 1 = x + 18

Subtract x from both sides

x + 1 = 18

Subtract 1 from both sides

x = 17

QPR = 2x + 1

QPR = 2(17) + 1

QPR = 35º

-------------------------------

8y - 4 = 4y + 28

Subtract 4y from both sides

4y - 4 = 28

Add 4 to both sides

4y = 32

Divide both sides by 4

y = 8

QPR = 8y - 4

QRP = 8(8) - 4

QRP = 60º

7 0

3 years ago

The temperature in the ice rink must stay below 50°F this morning the temperature was 71 Degrees Fahrenheit the icing runs a coo

ahrayia [7]

Answer:

The temperature needs to decrease 71° - 50° = 21° that at least not greater than 50°F

As we know, the device decrease the temperature down 3.5° every hour, so

21 : 3.5 = 6 hour

Call x is the hours we need to wait to let the device decrease, we have

x > 6

3 0

3 years ago

Select all the true statements.

dangina [55]

Answer:

Dont get it either im stuck on it

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Vince added 0.912,0.882 ,and 1.354 liters of liquid to a bottle.About how much is in the bottle?

Levart [38]

3.148 liters of liquid.

3 0

3 years ago

Find the volume of the wedge-shaped region contained in the cylinder x2 + y2 = 49, bounded above by the plane z = x and below by

fiasKO [112]

$\displaystyle\iiint_R\mathrm dV=\int_{y=-7}^{y=7}\int_{x=-\sqrt{49-y^2}}^{x=0}\int_{z=x}^{z=0}\mathrm dz\,\mathrm dx\,\mathrm dy$

Converting to cylindrical coordinates, the integral is equivalent to

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}\int_{z=r\cos\theta}^{z=0}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}-r^2\cos\theta\,\mathrm dr\,\mathrm d\theta$
$=-\displaystyle\left(\int_{\theta=\pi/2}^{3\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{r=0}^{r=7}r^2\,\mathrm dr\right)$
$=\dfrac{2\times7^3}3=\dfrac{686}3$

4 0

3 years ago