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tia_tia [17]
3 years ago
13

Gerard bought 9 hamburgers and 3 orders of fries for 24.75. Chris bought 6 hamburgers and 4 orders of fries for 19.50. Each hamb

urger cost the same amount. Each Order of fries cost the same amount. Wrote a system of equations that can be used to find how much one hamburger and one Order of fries cost.
Mathematics
1 answer:
dezoksy [38]3 years ago
8 0
<span>9 hamburgers and 3 fries for 24.75 and 6 hamburgers and 4 fries for 19.50 can be represented by 9h + 3f = 24.75 and 6h + 4f = 19.5 Multiply by 4 36h + 12f = 99 and multiply by -3 -18h -12f = -58.5 12f and -12f cancel each other out 36h - 18h = 99 -58.5 18h = 40.5 h = 2.25 hamburgers cost 2.25 9h + 3f = 24.75 9(2.25) + 3f = 24.75 3f = 24.75 - 20.25 = 4.50 f = 150 fries cost 1.50 One hamburger and one fries can be represented by : h + f 2.25 + 150 = 3.75</span>
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The value of h(t) when t=\frac{15}{32} is 10.02.

Solution:

Given function h(t)=-16t^2+15t+6.5

To find the value of h(t) when t=\frac{15}{32}:

h(t)=-16t^2+15t+6.5

Substitute t=\frac{15}{32} in the given function.

$h\left(\frac{15}{32} \right)=-16\left(\frac{15}{32} \right)^2+15\left(\frac{15}{32} \right)+6.5

            $=-16\left(\frac{225}{1024} \right)+15\left(\frac{15}{32} \right)+6.5

Now multiply the common terms into inside the bracket.

           $=-\left(\frac{3600}{1024} \right)+\left(\frac{225}{32} \right)+6.5

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          $=-\left(\frac{225}{64} \right)+\left(\frac{225}{32} \right)+6.5

To make the denominator same, take LCM of the denominators.

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        $=-\left(\frac{225}{64} \right)+\left(\frac{225\times2}{32\times2} \right)+6.5\times\frac{64}{64}

        $=-\frac{225}{64} +\frac{450}{64}+\frac{416}{64}

        $=\frac{-225+450+416}{64}

       $=\frac{641}{64}

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$h\left(\frac{15}{32} \right)=10.02

Hence the value of h(t) when t=\frac{15}{32} is 10.02.

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3 years ago
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