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erik [133]
2 years ago
7

How do you find the area of the shaded sector of the circle? The radius is 10in and the sector is 45 degrees

Mathematics
1 answer:
Step2247 [10]2 years ago
7 0

Given:

Radius of the circle = 10 in

Central angle of the sector = 45 degrees

To find:

The area of the sector.

Solution:

Area of a sector is

A=\dfrac{\theta}{360^\circ}\pi r^2

Where, \theta is the central angle in degrees.

Putting r=10 and \theta=45^\circ, we get

A=\dfrac{45^\circ}{360^\circ}\pi (10)^2

A=\dfrac{1}{8}\pi (100)

A=12.5\pi

Therefore, the area of the sector is 12.5π sq. inches.

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Expand and simplify (x+3) (x-4)
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Answer:

Step-by-step explanation:

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Gilbert is graduating from college in one year, but he will need a loan in the amount of $5,125 for his last two semesters. He m
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First, determine the effective interests given both interest rates.

(1)   ieff = (1 + 0.068/12)^12 - 1 = 0.07016
(2)   ieff = (1 + 0.078/12)^12 - 1 = 0.08085

Calculating the interests will entail us to use the equation,
     I = P ((1 + i)^n - 1)

Substituting the known values,
  (1)    I = ($5125)((1 + 0.07016)^1/2 - 1)
          I = $176.737

   (2)    I = ($5125)(1 + 0.08085)^1/2 - 1)
            I = $203.15

a. Hence, the greater interest will be that of the second loan. 

b. The difference between the interests,
    d = $203.15 - $176.737
          $26.413
7 0
3 years ago
If 3x2 - 2x - 16 represents the area of a rectangle, what is the perimeter of that rectangle?
zubka84 [21]
The answer should be B if I am correct
4 0
2 years ago
three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

7 0
3 years ago
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