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kramer
4 years ago
7

In the square below, the two semi-circles are congruent. Find the area of the shaded region. If necessary, round your answer to

two decimal places.
the square side measures at 8in

Mathematics
1 answer:
Jobisdone [24]4 years ago
3 0
Let's start with the area of the square
area \: square = s \times s = 8 \times 8 = 64
now let's subtract the are of the two half circles.
two half circles are the same as one circle, and we know that the diameter of the circle is 8 (same as the side of a square) so it's radius is 8/2= 4 inches

area \: circle = \pi {r}^{2}  = \pi \times  {4}^{2}  = 16\pi
now we just subtract and our answer is

64 - 16\pi = 64 - 16 \times (3.14) = 13.73
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Step-by-step explanation:

The law of cosines tells you ...

  b² = a² +c² -2ac·cos(B)

Substituting for a²+c² using the given equation, we have ...

  b² = b²·cos(B)² -2ac·cos(B)

We can subtract b² to get a quadratic in standard form for cos(B).

  b²·cos(B)² -2ac·cos(B) -b² = 0

Solving this using the quadratic formula gives ...

  \cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...

  \cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

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-2/3 is the simplified form of the expression -1/2 ÷ 3/4.

<h3>What is the simplified form of the given expression?</h3>

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To divide by a fraction, multiply by its reciprocal.

-1/2 ÷ 3/4

Reciprocal of 3/4 is 4/3

Hence

-1/2 × 4/3

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( -1 × 2 ) / ( 1 × 3 )

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1 year ago
A Japanese bullet train travels 558 miles per 3 hours A. What is the rate B. How far does the train travel in 1 hour WHO EVER A
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Answer:

Step-by-step explanation:

Remark

The rate is going to be the same as the distance travelled in 1 hour. The units will be different.

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Problem A

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Problem B

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Solution

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This looks really trivial, but it's not. You have to learn to see the difference between a number and its units. It's not very often that the numbers will be the same, but if the units differ, then it is an entirely different question.

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