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morpeh [17]
2 years ago
12

Which expression results in a sum or difference of five and six fifty fourths? (3 points) a eight and seven ninths minus three a

nd four sixths b six and eight ninths minus one and five sixths c four and one sixths plus one and one ninth d three and three sixths plus two and four ninths
Mathematics
1 answer:
Genrish500 [490]2 years ago
6 0

Answer:

a)  eight and seven ninths minus three and four sixths

Step-by-step explanation:

The required result is 5\frac{6}{54}

a)

8\frac{7}{9} -3\frac{4}{6}\\\\=\frac{79}{9}  -\frac{22}{6} \\\\Simplifying\ the\ expression:\\\\=\frac{474-198}{54} \\\\=\frac{276}{54}\\\\=5\frac{6}{54}

b)

6\frac{8}{9} -1\frac{5}{6}\\\\=\frac{62}{9}  -\frac{11}{6} \\\\Simplifying\ the\ expression:\\\\=\frac{372-99}{54} \\\\=\frac{273}{54}\\\\=5\frac{3}{54}

c)

4\frac{1}{6} +1\frac{1}{9}\\\\=\frac{10}{9}  +\frac{25}{6} \\\\Simplifying\ the\ expression:\\\\=\frac{60+225}{54} \\\\=\frac{285}{54}\\\\=5\frac{15 }{54}

d)

3\frac{3}{6} +2\frac{4}{9}\\\\=\frac{22}{9}  +\frac{21}{6} \\\\Simplifying\ the\ expression:\\\\=\frac{132+189}{54} \\\\=\frac{321}{54}\\\\=5\frac{51 }{54}

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serious [3.7K]
Well hmmm let's say you take the car and go in the city for 60 miles with it, well, the car can do 60 miles per gallon, since you just drove it for 60 miles, you only spent 1 gallon of gasoline then.

that only happens if you drive it for 60 miles, what if you drive it for more, let's do a quick table on that,

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
60&3.6(1)\\\\
&3.6\left( \frac{60}{60} \right)\\\\
120&3.6(2)\\\\
&3.6\left( \frac{120}{60} \right)\\\\
180&3.6(3)\\\\
&3.6\left( \frac{180}{60} \right)
\end{array}

and so on, now let's check if you less than 60 miles,

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
40&3.6\left( \frac{40}{60} \right)\\\\
20&3.6\left( \frac{20}{60} \right)\\\\
10&3.6\left( \frac{10}{60} \right)
\end{array}

so, if you divide the amount of miles driven, by 60, when you have driven it for 120 miles, 120/60 is just 2, and the cost is for 2 gallons, or 3.6 * 2, which is 7.2 bucks, for 180 miles is 180/60 or 3 gallons for 3.6 * 3 bucks, and so on.

now, what if you drive it instead for "m" miles?

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
m&3.6\left( \frac{m}{60} \right)\\\\
\end{array}\implies c=3.6\left( \cfrac{m}{60} \right)\implies c=\cfrac{3.6m}{60}
\\\\\\
c=\cfrac{3.6}{60}m\implies c=0.06m
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