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Delicious77 [7]
3 years ago
12

Keisha's parents want to save twenty thousand dollars in her college savings account over the next fifteen years. They have eigh

t thousand dollars to use as an initial deposit. What simple annual interest rate do they need to meet their goal? Round your answer to the nearest tenth. Select one: A. two point five percent B. two point seven percent C. ten percent D. twelve percent
Mathematics
1 answer:
Phantasy [73]3 years ago
6 0

The correct answer is C because if you use the simple interest formula I=PRT The interest has to be $8000+X=$20,000 (X=12,000) P stands for the beginning amount of money ($8000) the R stands for the rate (you don't know this for sure yet) and the T stands for the amount of time(15 years) $12,000=$8000(R)(15) $8000(15)=120,000 $12,000=120,000(R)  $12,000/120,000=120,000(R)/120,000 R=0.1 then multiply by 100 to get the percent of 10% therefore C is your answer.
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Answer:

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Step-by-step explanation:

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c = b / cos 45.

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3 years ago
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Luba_88 [7]

Answer:

11 mph and 20 mph

Step-by-step explanation:

Represent his average speed going by r1 and his average speed returning by r2.  We know that r1 = r2 + 9.

Recall that distance = rate times time, so time = distance / rate.

Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).

Time spend returning was (280 mi) / r2.

The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs

Note that there is only one variable here:  r2.  Find r2, and then from r2, find r1:

Dividing all 3 terms by 14 hrs yields:

  20            20

---------- + ----------- = 1

r2 + 9         r2

The LCD here is r2(r2 + 9).  Thus, we have:

      20r2                    (r2 +  9)(r2)

------------------- = 1 or  ------------------

 (r2 +  9)(r2)               (r2 +  9)(r2)

Then 20(r2) = (r2)^2 + 9(r2).  This is reducible by dividing all terms by r2:

20 = r2 + 9, or 11 = r2.  Then r1 = 11 + 9, or 20.

The two rates were 11 mph and 20 mph.

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blep

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4 years ago
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Two main facts are needed here:

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By (2) we know that \sqrt{2x-1} exists if 2x-1\ge0, or x\ge\dfrac12.

By (1), we know that \log(\sqrt{2x-1}+3) exists if \sqrt{2x-1}+3>0, or \sqrt{2x-1}>-3. But as long as the square root exists, it will always be positive, so this condition will always be met.

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