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mixas84 [53]
3 years ago
10

Round the number to the hundredths place.

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0
First, look at the digits less than one one-hundredth (\frac{1}{100}, or .01):
0.00416
See if that number is closer to the higher or lower hundredth:
(if the first digit is 5 or greater, then higher;
if the first digit is 4 or lower, then lower)
Since 4 is closer to the lower hundred, round down
0.00416 -> 0.00000
Add this to the digits we didn't touch:
<span>17,546.32000 + 0.00000 = 17,546.32
</span>17,546.32416 rounded to the nearest hundredth is C. <span>17,546.32</span>
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1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

Steps:

\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}

\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

Steps:

\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
3 years ago
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