What is the domain of the function shown in the graph below?

Write the equation of the line that is parallel to the line y= -2x+3 and goes through the point (3, -6).

svetlana [45]

Answer:

y=-2

Step-by-step explanation:

because it's parallel --> a=-2

therefore we have y=-2x+b

the line goes through the point (3,-6) --> (-2).3+b=-6 --> b=0

3 0

3 years ago

Give me a real answer or else

ElenaW [278]

Answer:

1. scholars team is 15 %, student gov. 0.08%, band 24%, sports 34 %, fun raising is 17%

2. athletics is 30%, gov is 0.07 %, band is 21%, fund raising is 15%, scholars team is 14%

Step-by-step explanation:

7 0

3 years ago

What is the formula for Inflation Rate?

snow_lady [41]

Inflation Rate= CPI2-CPI1/CP1 x 100

3 0

3 years ago

Michael drove from home to work at an average speed of 45 mph. His return trip home took 45 minutes longer because he could only

Xelga [282]

The distance from his home to his work place is 118.125 miles

<h3>How to find the distance from his home to his work place?</h3>

He drove from home to work at an average speed of 45 mph. Therefore,

speed = 45 mph

let

time = x

speed = distance / time

distance = 45x

His return trip home took 45 minutes longer because he could only drive at 35 mph. Therefore,

45 minutes = 0.75 hours

distance = 35(x + 0.75)

distance = 35x + 26.25

Therefore,

45x = 35x + 26.25

45x - 35x = 26.25

10x = 26.25

x = 26.25 / 10

x = 2.625

Therefore,

distance = 45(2.625) = 118.125 miles

learn more on distance here: brainly.com/question/21329221

#SPJ1

3 0

2 years ago

Please help? I’m super lost...

babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function. Watch and see.

a. $sin2\theta =-\frac{\sqrt{3} }{2}$

Take the inverse sin of both sides. When you do that, you are left with just 2theta on the left. That's why you do this.

$sin^{-1}(sin2\theta)=sin^{-1}(-\frac{\sqrt{3} }{2} )$

This simplifies to

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

We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2. Those are:

$2\theta =\frac{5\pi }{6}$ and

$2\theta=\frac{7\pi }{6}$

Divide both sides by 2 in both of those equations to get that values of theta are:

$\theta=\frac{5\pi }{12},\frac{7\pi }{12}$

b. $tan(7a)=1$

Take the inverse tangent of both sides:

$tan^{-1}(tan(7a))=tan^{-1}(1)$

Taking the inverse tangent of the tangent on the left leaves us with just 7a. This simplifies to

$7a=tan^{-1}(1)$

We look to the unit circle to find which values of <em>a</em> give us a tangent of 1. They are:

$7\alpha =\frac{5\pi }{4},7\alpha =\frac{\pi }{4}$

Dibide each of those equations by 7 to find that the values of alpha are:

$\alpha =\frac{5\pi}{28},\frac{\pi}{28}$

c. $cos(3\beta)=\frac{1}{2}$

Take the inverse cosine of each side. The inverse cosine and cosine undo each other, leaving us with just 3beta on the left, just like in the previous problems. That simplifies to:

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

We look to the unit circle to find the values of beta that give us the cosine of 1/2 and those are:

$3\beta =\frac{\pi}{6},3\beta =\frac{5\pi}{6}$

Divide each of those by 3 to find the values of beta are:

$\beta =\frac{\pi }{18} ,\frac{5\pi}{18}$

d. $sec3\alpha =-2$

Let's rewrite this in terms of a trig ratio that we are a bit more familiar with:

$\frac{1}{cos(3\alpha) } =\frac{-2}{1}$

We are going to simplify this even further by flipping both fraction upside down to make it easier to solve:

$cos(3\alpha)=-\frac{1}{2}$

Now we will take the inverse cos of each side (same as above):

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

We look to the unit circle one last time to find the values of alpha that give us a cosine of -1/2:

$3\alpha =\frac{7\pi}{6},3\alpha =\frac{11\pi}{6}$

Dividing both of those equations by 3 gives us

$\alpha =\frac{7\pi}{18},\frac{11\pi}{18}$

And we're done!!!

8 0

2 years ago