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Brrunno [24]
3 years ago
6

For a particular cable,the radius of the base of the inner layer is 2 millimeters (mm) and the radius of the base of the outer l

ayer is 6 mm.What is the percent of the area of the base pf the outer layer is the area of the inner layer?
Mathematics
1 answer:
andrew11 [14]3 years ago
4 0

Answer:

Step-by-step explanation:

The formula for determining the area of a circle is expressed as

Area = πr²

Where

π is a constant whose value is 3.14

r represents the radius of the circle.

Considering the inner layer

radius = 2 millimeters

Area of base = 3.14 × 2² = 12.56 mm²

Considering the outer layer,

Radius = 6 mm

Area of base = 3.14 × 6² = 113.04 mm²

Percent of the area of the base of the outer layer that is the area of the base of the inner layer is

12.56/113.04 × 100 = 11.1%

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Which is a factor of 10 but not a mutiple of 2? choices are 8,5,4,and 6
Verizon [17]

5 is Your answer,5 factors into 10 but none of the other choices do


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strojnjashka [21]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
A bag contains 5 blue marbles, 2 green marbles, and 3 yellow marbles. What is the probability of choosing one green marble and t
Vesna [10]

Answer:

Step-by-step explanation:

Total number of marbles = 5 + 2 + 3 = 10

Probability of choosing 1 green marble = 2/10

Probability of choosing 1 yellow marble = 3/10

Notice (and this is important) that the denominator didn't change. Why?

Because you replaced the first marble into the bag. That word replacement is critical in a problem of this nature. There is the term non replacement which means that the second draw would have a denominator of 9.

So what is the probability of P(green then yellow)?

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4 0
2 years ago
Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m
andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

   A_1  =  67.58 \ in^2

2

   A_2 =415.4 \ ft^2

3

   A_3  =  8.48 \ cm^2

4

  A_4 =  480.38 \ m^2

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

         A =  \frac{\theta}{360} * \pi r^2

Now at r_1  = 11 in and  \theta_1 =  64^o

       A_1 =  \frac{64}{360} * 3.142  * 11^2

       A_1  =  67.58 \ in^2

Now at  r_2  = 20 ft in and  \theta_2  =  119 ^o

       A_2 =  \frac{119}{360} * 3.142 *  20^2

       A_2 =415.4 \ ft^2

Now at  r_3  = 6.5 cm   and  \theta_3 =  23 ^o

     A_3  =  \frac{23}{360} * 3.142 *  6.5 ^2

      A_3  =  8.48 \ cm^2

Now at  r_4  = 14.2 m   and  \theta_4 = 360 -87 =  273 ^o

         A_4 =  \frac{273}{360}  * 3.142 * 14.2^2

          A_4 =  480.38 \ m^2

6 0
3 years ago
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