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2 years ago

9

In the square below the four quarter circles are congruent. find the area of the shaded region.

1 answer:

Alexus [3.1K]2 years ago

5 0

The quarter circle pieces are all congruent allowing us to glue them together to form a circle. The circle will be completely inside the square. This circle is as large as possible so it's not spilling outside of the square.

The area of the square is 8*8 = 64 square units

The circle has a radius of 8/2 = 4 units. Use r = 4 and plug this into the area of a circle formula
A = pi*r^2
A = pi*4^2
A = pi*16
A = 16pi

Now subtract the two areas
(area of square) - (area of circle) = 64 - 16pi

The exact answer, in terms of pi, is 64 - 16pi

If your teacher wants you to find the approximate decimal version, then you would use a calculator to find that 64 - 16pi = 13.7345175 approximately. If you end up going with the approximate answer, be sure to follow the instructions on what your teacher wants with rounding.

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X + 2y = 5 3x + 5y = 14 Solve the system of equations. (3, 1) (7, -1) (2, 3/2)

aniked [119]

Answer:

The solution is (3 , 1)

Step-by-step explanation:

To solve the system of the equation you can use the elimination method or substitution method.

We will use the substitution method

From the first equation x + 2y = 5⇒get x = 5 - 2y

In the second equation 3x + 5y = 14⇒we will substitute x by 5 - 2y

3(5 - 2y) + 5y =14

15 - 6y + 5y = 14

-6y + 5y = 14 - 15

-y = -1

y = 1

Substitute the value of y in x = 5 - 2y

x = 5 - 2(1) = 5 -2 =3

∴ The solution of the system of equations is (3 , 1)

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3 years ago

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We have that

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the answer is
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2 years ago

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The 12th term of the arithmetic sequence is 10.5. The 18th term of this sequence is 13.5. Find the common difference and the fir

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Answer:

common difference is 0.5

first term is 5

Step-by-step explanation:

<em>use </em><em>the </em><em>formula</em><em> for</em><em> </em><em>the</em><em> </em><em>nth </em><em>term </em><em>of </em><em>an </em><em>ap </em><em>Tn=</em><em>a+</em><em>(</em><em>n-1)</em><em>d</em>

<em>T12=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>T18=</em><em>1</em><em>3</em><em>.</em><em>5</em>

<em>therefore</em><em> </em><em>come </em><em>up </em><em>with </em><em>two </em><em>equations</em>

<em>T12=</em><em>a+</em><em>(</em><em>1</em><em>2</em><em>-</em><em>1</em><em>)</em><em>d</em>

<em>1</em><em>0</em><em>.</em><em>5</em><em>=</em><em>a+</em><em>1</em><em>1</em><em>d</em><em>(</em><em>1</em><em>s</em><em>t</em><em> </em><em>equation</em><em>)</em>

<em>T18</em><em>=</em><em>a+</em><em>(</em><em>1</em><em>8</em><em>-</em><em>1</em><em>)</em><em>d</em>

<em>1</em><em>3</em><em>.</em><em>5</em><em>=</em><em>a+</em><em>1</em><em>7</em><em>d</em><em>(</em><em>2</em><em>n</em><em>d</em><em> </em><em>equation</em><em>)</em>

<em>then </em><em>solve </em><em>both </em><em>as </em><em>a </em><em>simultaneous</em><em> </em><em>equation</em>

<em>a+</em><em>1</em><em>1</em><em>d</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>a+</em><em>1</em><em>7</em><em>d</em><em>=</em><em>1</em><em>3</em><em>.</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em>-6d/</em><em>-</em><em>6</em><em>=</em><em>-</em><em>3</em><em>/</em><em>-</em><em>6</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d=</em><em>0</em><em>.</em><em>5</em>

<em>use </em><em>one </em><em>of</em><em> the</em><em> </em><em>equations</em><em> </em><em>to </em><em>find </em><em>the </em><em>first</em><em> </em><em>term</em>

<em>a+</em><em>1</em><em>1</em><em>(</em><em>0</em><em>.</em><em>5</em><em>)</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

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<em>a=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>-</em><em>5</em><em>.</em><em>5</em>

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<em>I </em><em>hope </em><em>this </em><em>helps</em>

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3 years ago

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3 years ago

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Each person in a group of college students was identified by graduating year and asked when he or she preferred taking classes:

swat32

Answer:

<em>0.615</em>

Step-by-step explanation:

The frequency table is attached below.

We have to calculate, the probability that the student preferred morning classes given he or she is a junior.

i.e $P(\text{Morning }|\text{ Junior})$

We know that,

$P(A\ |\ B)=\dfrac{P(A\ \cap\ B)}{P(B)}$

So,

$P(\text{Morning }|\text{ Junior})=\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}$

Putting the values from the table,

$\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}=\dfrac{\frac{16}{143}}{\frac{26}{143}}=\dfrac{16}{26}=0.615$

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3 years ago