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3 years ago

10

Write an equation for the line, in point-slope form, that passes through the following point and has the following slope:slope:-

4 point: (1, -2)

1 answer:

enyata [817]3 years ago

5 0

Answer:

y+2=-4(x-1)

Step-by-step explanation:

y-y1=slope(x-x1)

just plug in the numbers you were given and youre good to go

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HURRYYYYY !!!!! Examine the following expression. p squared minus 3 + 3 p minus 8 + p + p cubed Which statements about the expre

algol [13]

Answer:

The terms 3p and p are like terms.

The terms in the expression are p squared, negative 3, 3p, p, negative 8, p, p cubed.

The expression contains 6 terms.

The terms p squared and p cubed are like terms.

6 0

3 years ago

Age of car = 8 years. Original cost = $18,000. The cost of maintenance and repairs is $

creativ13 [48]

From the graph it seems to be around the area of 10%.

So,

18,000 * 0.10 = $1,800

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3 years ago

Shawna earns $10 per hour babysitting her neighbors. One Saturday she got Jill to assist her for 40% of everything she earns. Th

Nastasia [14]

Answer:10 dollars

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3 0

3 years ago

What is the fractional part of a ton is a pound?

zvonat [6]

What is a ton to start off?
And how many pounds are in a ton
The fractional part would be how many pounds are in the ton

7 0

3 years ago

Read 2 more answers

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

3 years ago