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zavuch27 [327]
3 years ago
9

Three identical circular coins are lined up in a row as shown (the 10s represent their value, not a length). The distance betwee

n the first and third coins is 3.2 cm. What is the diameter of one of these coins?​

Mathematics
1 answer:
son4ous [18]3 years ago
3 0

Answer:

Diameter of a circle = 1.6 Cm

Step-by-step explanation:

Given:

Distance between first and third coins = 3.2 cm

Find:

Diameter of the coins = ?​

Computation:

Number of circle = 3

In the given figure,distance between the first and third coins is 3.2 cm.

So,

Distance = 1st circle radius + 2nd circle diameter + 3rd circle radius

Distance = 1st circle radius + 2 (radius) + 3rd circle radius

Distance = R + 2R + R

3.2 Cm = 4 R

Radius = 0.8 Cm

Diameter of a circle = 2 × radius

Diameter of a circle = 2 × 0.8

Diameter of a circle = 1.6 Cm

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Step-by-step explanation:

the ways to detect the hypotenuse of a triangle are :

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He normal pulse rate of a 13-year-old is between 70 and 100 beats per minute. As part of a health program, the medical staff at
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From the box plot, it can be seen that for grade 7 students,
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Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10

</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The least value is 76 and the highest value is 97. The lower and the upper quartiles are 85 and 94 respectively while the median is 89.

Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9

The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5

Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
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3 years ago
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3 years ago
What the answer to 30-32
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30. C & D

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3 years ago
What is the area of △FGH to the nearest tenth of a square meter? The image is of a triangle GHF with base GH length 2m, FG is 2.
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First, we are going to use the law of cosines to find the length of the line segment FH:
FH= \sqrt{2.5^{2}+2^{2}-(2)(2.5)Cos(121)}
FH= \sqrt{2.5^{2}+2^{2}-5Cos(121)}
FH=3.2

Next, we are going to use the semi-perimeter formula: s= \frac{GH+FG+FH}{2}
s= \frac{2+2.5+3.2}{2}
s= \frac{7.7}{2}
s=3.9

Now that we have the semi-perimeter of our triangle, we can find its area using Heron's formula:
A= \sqrt{s(s-GH)(s-FG)(s-FH)}
A= \sqrt{3.9(3.9-2)(3.9-2.5)(3.9-3.2)}
A= \sqrt{3.9(1.9)(1.4)(0.7)}
A=2.7m^{2}

We can conclude that the area of the triangle <span>GHF is 2.7 </span>m^{2}.

3 0
4 years ago
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