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kotegsom [21]
3 years ago
8

Can you help me solve this radical equation √y-10 = y-2 the sqrt sign is only over the y-10

Mathematics
1 answer:
matrenka [14]3 years ago
7 0
D:y-10\geq0 \wedge y-2\geq0\\
D:y\geq10 \wedge y\geq2\\
D:y\geq10\\\\
\sqrt{y-10}=y-2\\
y-10=(y-2)^2\\
y-10=y^2-4y+4\\
y^2-5y+14=0\\
\Delta=(-5)^2-4\cdot1\cdot14=25-56=-31\\
\Delta
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I need help plz you leave you house at 1 p.m to go to a wedding. The ceremony starts at 5 p.m. and is 350 miles away. You drive
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You leave at 1 P.M., the wedding ceremony starts at 5 P.M.

This gives you a 4 hour gap, the Wedding is 350 miles away, and you're traveling at 65 miles an hour.

Simply you can use two methods, (65 * 4) to verify if you'd make it or not or (350/65) to ensure how many hours it would take in total.

65 * 4 = 260 | Since this is not 350, we can verify that you'd be late.
350/65 = 5.4 (estimated) | Meaning it'd take about 5.4 hours.

To know how long that is we'd have to convert over.

5.4 = 5 2/5ths, 60 minutes are in an hour. Divide 60 by 5 to get 12, meaning:
1/5ths = 12/60ths, with that said multiply that by 2 to get your answer.
2/5ths = 24/60ths or 24 minutes.

This means you'd take 5 hours and 24 minutes to get there.

Originally we stated that the process would take 4 hours. Subtract:

(5 hours and 24 minutes) - (4 hours) = (1 hour and 24 minutes)

Concluding that you'd be late by an hour and 24 minutes.

I hope this helps, have a great rest of your day! ^ ^
| | Ghostgate | |
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