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jonny [76]
3 years ago
15

Only 3 students in each event win medals at the track meet. If 9 students are running the mile, what fraction of them will win a

medal?
Mathematics
2 answers:
jeyben [28]3 years ago
8 0
The other person said 3/9. While this is correct, you have to simplify the fraction. Both numbers are divisible by 3... 3/3=1 and 9/3=3. So, the answer would be 1/3.
Umnica [9.8K]3 years ago
4 0
I think the answer is 3/9 because there is 9 students running but only 3 of them win medals so I think it would be 3/9 or maybe something else
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3 0
3 years ago
Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.
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Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152

To find the area of the circumscribed circle:

r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

AC=\sqrt{10.152^2-8^2}\approx 6.25

The area of the triangle is:

A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25

The semiperimeter of the triangle is:

\dfrac{10.152+6.25+8}{2}\approx 24.4

The radius of the circle is therefore \dfrac{25}{24.4}\approx 1.025

The area of the inscribed circle then is \pi\cdot (1.025)^2\approx 3.298.

Hope this helps!

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Step-by-step explanation:

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Step-by-step explanation:

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