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anygoal [31]
4 years ago
10

A target has a bull’s-eye with a diameter of 2 inches. the outer ring is 1 inch wide. what is the area of the outer ring?

Mathematics
2 answers:
Ghella [55]4 years ago
4 0

Answer:

ITS A

Step-by-step explanation:

andrew11 [14]4 years ago
3 0
The diameter of the outer ring will be 4 inches since it extends one inch beyond the bull's eye from either side. See the attached image for clarification.

The area of the outer ring can be expressed as:
Area_{outerring} = Area_{outercircle} - Area_{bullseye}
This is because we need to take into account of the space taken up by the bullseye. 

Now we set up our equation:
Area_{outerring} =  \pi r^{2}_{outercircle} -  \pi r^{2}_{bullseye}

Now plug and solve:
Area_{outerring} =  \pi 2^{2} -  \pi 1^{2}
Area_{outerring} =  4\pi -  \pi
Area_{outerring} = 9.42477796

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Can you explain to me what is 20% of 150, and how to get it?
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Answer:

30

Step-by-step explanation:

20% = 20/100= 1/5

It is 1/5 of the number 150

Taking the problem what is 20% of 150

Is means equals and of means multiply

what is 20% of 150

W = 20% * 150

Change 20% to a decimal

W = .20 * 150

W = 30

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3 years ago
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Step-by-step explanation:

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3 years ago
Solve by factoring:<br><br> 25X^2+5X-12=0
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3 0
4 years ago
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Evaluate the indicated limit algebraically. Change the form of the function where necessary. Please write clearly with descripti
alukav5142 [94]

Answer:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3

Step-by-step explanation:

We want to evaluate the limit:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}

To do so, we can divide everything by <em>x</em>². So:

=\displaystyle \lim_{x\to \infty}\frac{3+4.5/x^2}{1-1.5/x^2}

Now, we can apply direct substitution:

\Rightarrow \displaystyle \frac{3+4.5/(\infty)^2}{1-1.5/(\infty)^2}

Any constant value over infinity tends towards 0. Therefore:

\displaystyle =\frac{3+0}{1+0}=\frac{3}{1}=3

Hence:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3

Alternatively, we can simply consider the biggest term of the numerator and the denominator. The term with the strongest influence in the numerator is 3<em>x</em>²<em>, </em>and in the denominator it is <em>x</em>². So:

\displaystyle \Rightarrow \lim_{x\to\infty}\frac{3x^2}{x^2}

Simplify:

\displaystyle =\lim_{x\to\infty}3=3

The limit of a constant is simply the constant.

We acquire the same answer.

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Answer:

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Step-by-step explanation:

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