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GarryVolchara [31]
3 years ago
5

At 3 P.M., the temperature is 12°F. It then began dropping 2°F every hour for 7 hours. What was the temperature after 7 hours?

Mathematics
1 answer:
fiasKO [112]3 years ago
7 0
All you need to do to answer this question is multiply 2 (the dropping temperature) by 7 (time in hours) which is 14. Subtract 14 from 12 and your answer is -2 degrees Fahrenheit
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Simplify the radical expression completely LaTeX: 3\sqrt{5}\:-\:3\sqrt{11}+2\sqrt{121}-3\sqrt{90}3 5 − 3 11 + 2 121 − 3 90 A. La
lora16 [44]

Answer:

3\sqrt{5}\:-\:3\sqrt{11}+22-9\sqrt{10}

Step-by-step explanation:

Given

3\sqrt{5}\:-\:3\sqrt{11}+2\sqrt{121}-3\sqrt{90}

Required

Simplify the expression

3\sqrt{5}\:-\:3\sqrt{11}+2\sqrt{121}-3\sqrt{90}

Express the square root of 121 as 11

3\sqrt{5}\:-\:3\sqrt{11}+2 * 11-3\sqrt{90}

3\sqrt{5}\:-\:3\sqrt{11}+22-3\sqrt{90}

Express 90 as 9 * 10

3\sqrt{5}\:-\:3\sqrt{11}+22-3\sqrt{9 * 10}

Split the surd to 2

3\sqrt{5}\:-\:3\sqrt{11}+22-3\sqrt{9} * \sqrt10}

Express the square root of 9 as 3

3\sqrt{5}\:-\:3\sqrt{11}+22-3 * 3 * \sqrt10}

3\sqrt{5}\:-\:3\sqrt{11}+22-9 * \sqrt10}

3\sqrt{5}\:-\:3\sqrt{11}+22-9\sqrt{10}

Hence, the result of the expression is

3\sqrt{5}\:-\:3\sqrt{11}+22-9\sqrt{10}

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poizon [28]

9514 1404 393

Answer:

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Step-by-step explanation:

It can help to give all the ratios a common denominator. (We'll use 100, and fudge the last one a bit.)

  8/10 = 80/100 . . . . our reference value

__

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  3/5 = 60/100 . . . . less than 8/10

  50/100 . . . . . . . . . less than 8/10

  13/15 ≈ 86.7/100 . . . more than 8/10

6 0
3 years ago
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