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nekit [7.7K]
3 years ago
11

Help finding the area

Mathematics
1 answer:
snow_lady [41]3 years ago
6 0

Answer:

The answer is 49 in^2.

Step-by-step explanation:

The area of the triangle is 1/2* perpendicular height * horizontal length.

The horizontal length is 14 in, while the perpendicular height is 7 in.

The area of the triangle is 1/2* 7 in * 14 in = 49 in^2

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400mm^2

Step-by-step explanation:

8 0
3 years ago
Write 77/20 as a terminating decimal
ahrayia [7]
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What's the answer and how to get it
Murrr4er [49]
The answer is B.)
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8 0
3 years ago
For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
3 0
3 years ago
Find the gradients of lines A and B.
kherson [118]
Hey, you didn’t provide any picture or further context.

To solve for gradient, do (y2-y1)/(x2-x1)
So what do you think?
5 0
3 years ago
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