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Ierofanga [76]
3 years ago
10

Ursula and Andre competed against each other in a swimming race. Ursula swam at a rate of 60 meters per minute and Andre swam at

a rate of 48 meters per minute. Ursula gave Andre a 1-minute head start.
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0
Given that the two swimmers competed and Ursula's speed is 60 m/min while Andre's speed is 48 m/min. The distance that the Ursula will catch up with Andre will be:
distance=(relative speed)×(time)
relative speed=60-48=12 m/min
the two swimmers met at a distance of:
12×1
=12 meters

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6 0
3 years ago
Read 2 more answers
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lapo4ka [179]
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3 years ago
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Vinvika [58]
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6 0
3 years ago
Need help with 9 and 12(theirs a picture)
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8 0
3 years ago
I need help thank you
erma4kov [3.2K]

Answer:

≈ 5

Step-by-step explanation:

<u><em>You can use the Pythagorean Theorem to solve this:</em></u>

a^{2} + b^{2} = c^{2}

<u><em>Now plug in the numbers into the formula:</em></u>

6^{2} + b^{2} = 8^{2}

36 + b^{2} = 64

<u><em>Subtract 36 from both sides:</em></u>

36 + b^{2} = 64

-36        -36

_________

b^{2} = 28

<em><u>Square both sides:</u></em>

\sqrt{b^2} = \sqrt{28}

b = 5.29

b ≈ 5

4 0
3 years ago
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