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2 years ago

Find the slope of the line on the graph. Write the answer as a fraction or a whole number, not a mixed number or decimal.

Mathematics

1 answer:

Inga [223]2 years ago

6 0

(-3,0) and (0, 3)
slope = 3/3 = 1

answer
slope = 1

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A map of a park is created on a coordinate plane. Each until on the map represents 1 kilometer. A tennis court is located at (0,

pogonyaev

Answer:

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Step-by-step explanation:

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2 years ago

Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0

gtnhenbr [62]

Answer:

Step-by-step explanation:

Given is a differential equation as

$ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0$

Divide this by t to get in linear form

$y'+6y/t = t-1+1/t$

This is of the form

y' +p(t) y = Q(t)

where p(t) = 1/t $e^(\int 1/tdt) = t$

So solution would be

$yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C$

siubstitute y(1) = 16

$16 = 16^3/3-128+1+C\\C = -1206$

4 0

3 years ago

Convert 4 5/7 to an improper fraction

Musya8 [376]

Answer:

33/7

Step-by-step explanation:

8 0

3 years ago

GIVING BRAINLIEST FOR BEST ANSWER!

Stels [109]

Answer: The answer is 11 1/24 or in decimal form 11.04

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Someone help please

Alla [95]

Answer: Choice A

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

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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