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ra1l [238]
3 years ago
9

A ruler is 12 inches long. What is the length of this ruler in centimeters? Round to the nearest hundredth.

Mathematics
1 answer:
agasfer [191]3 years ago
4 0
I do not know but i can give you about 2 - 3 tips.
First get a ruler and figure out how many centimeters are in a inch.
Then you will multiply 12 and how many centimeters are in a inch.
And that is how you will get your answer.
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<span>Prove that AC and BD have the same midpoint. The diagonals of a parallelogram bisect each other; therefore, they have the same midpoint.</span>
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A rental car company chargea a flat daily fee plus a charge for each mile driven. A car rented for 5 days and driven for 300 mil
stellarik [79]
5x + 300y = 178
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3 0
3 years ago
3a+2x-3y=15<br>Solve for a.<br>Then Solve put the value for a =<br>a+7÷10​
erma4kov [3.2K]

\huge\bf Question:–

\sf \longmapsto \: 3a+2x−3y=15

\bf \huge \: To  \: Find:–

\boxed{\bf \: Value\: of  \: A}

\huge\bf Solution:–

\sf \longmapsto \: 3a+2x−3y=15

\boxed{ \bf \: Add -2x  \: to  \: both  \: sides}

\sf \longmapsto \: 3a+2x−3y+−2x=15+−2x

\sf \longmapsto \: 3a−3y=−2x+15

\boxed{ \bf \: Add  \: 3y  \: to \:  both  \: sides}

\sf \longmapsto \: 3a−3y+3y=−2x+15+3y

\sf \longmapsto \: 3a=−2x+3y+15

\boxed{\bf \:  \: Divide  \: both  \: sides \:  by \:  3}

\sf \longmapsto \:  \dfrac{3a}{3} =  \dfrac{−2x+3y+15}{3}

\boxed{\bf \:  Cross \: Multiply}

\boxed{\sf \longmapsto \: a =  \dfrac{ - 2}{3} x + y + 5}

______________________________________

\bf \: Put\:The\: Value

\sf \longmapsto \: \dfrac{−2x+3y+15}{3}  +7÷10

\boxed{\bf \: Distribute}

\sf \longmapsto \: \dfrac{ - 2}{3} x+y+5+ \dfrac{7}{10}

\boxed{\bf \: Combine \:  Like \:  terms}

\sf \longmapsto \: \bigg( \dfrac{ - 2}{3} x\bigg) + y +\bigg( 5 +  \dfrac{7}{10} \bigg)

\sf \longmapsto \:  \dfrac{ - 2}{3} x + y +  \dfrac{57}{10}

______________________________________

\boxed{\bf The Answer\: is:–}

\boxed{{\underline{\bf\dfrac{ - 2}{3} x + y +  \dfrac{57}{10}} }}

8 0
3 years ago
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10 1/2 I am pretty sure its right
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3 years ago
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