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babymother [125]
3 years ago
12

Find the equation of the line That is perpendicular to y=-2x+1 and contains the point (8,2)

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
5 0

Answer:

y=.5x-2

Step-by-step explanation:

If one line is perpendicular to another that means that their slopes are negative reciprocals

which means that the slope (or m in y=mx+b) is .5

so far we have

y=.5x+b

to solve for b we plug in (8,2)

2=.5(8)+b

2=4+b

-2=b

therefore our equation is

y=.5x-2

Flauer [41]3 years ago
3 0

Answer:

hope it helps

stay safe healthy and happy.

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Which is the equation of a line that has a slope of -2/3 and passes through point (-3, -1)? y=-2/3x+1.
bazaltina [42]

Answer:

y = -2/3x - 3

Step-by-step explanation:

First, write the equation using the slope. It should look like this: y = -2/3x + b. Next, find the y intercept by plugging in the point. This is what it looks like: -1 = -2/3(-3) + b. Multiply -2/3 by -3 to get 2. This is what it should look like now: -1 = 2 + b. Subtract 3 from both sides to get -3 = b. Go back to your original equation and plug in -3 for b. This is your final equation:

y = -2/3x - 3

Hope this helped! :)

7 0
3 years ago
Amber has some state quarters in her pocket. She collects the following data by randomly pulling one quarter recordings the stat
ch4aika [34]
To do this problem, we need one more piece of information. He need to know the percent of the quarters that are from New York.

If you had that number, just multiply it by 300.

For example, if 10% of the coins were from New York, just multiply by 0.1.
0.1 x 300 = 30
4 0
3 years ago
F(5)-f(3) = x^2+7x-22
My name is Ann [436]

Answer:

your answer is going to be -158

Step-by-step explanation:

(5)-(3)^2+7×-22=-158

4 0
2 years ago
True or false: the standard deviation is the square root of the variance.
Vilka [71]
Why not look up "standard deviation" yourself, to be sure?  But yes, the std. dev. is the sqrt of the variance.
6 0
3 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
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