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ANTONII [103]
3 years ago
10

You own a lawn care business. in a typical week you drive 600 miles and use 40 gallons of gasoline. gasoline costs $1.25 per gal

lon, and your truck's fuel effiency is 21 miles per gallon on the highway and 13 miles per gallon in town. how much money do you spend on gasoline to travel to out-of-town clients?
Mathematics
1 answer:
iVinArrow [24]3 years ago
4 0
42.50$ ?        21 times 1.25  plus 13 times 1.25 = 42.50?  
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lilavasa [31]

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$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}

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The simplified fraction.

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Step 1: Simplify the numerator

$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}

Step 2: Simplify the denominator

$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}

Step 3: Using step 1 and step 2

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}

Step 4: Using fraction rule:

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}

$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}

Cancel the common factor r and t², we get

           $=\frac{64 r^{2} \cdot 9 }{16  \cdot 15 t^2 }

Cancel the common factors 16 and 3 on both numerator and denominator.

           $=\frac{4 r^{2} \cdot 3 }{  5 t^2 }

           $=\frac{12 r^{2}  }{  5 t^2 }

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2}  }{  5 t^2 }

The simplified fraction is \frac{12 r^{2}  }{  5 t^2 }.

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