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den301095 [7]
3 years ago
15

Michelle decides to mix grades of gasoline in her truck. She puts in 9 gallons of regular and 11 gallons of premium for a total

cost of $62.38. If premium gasoline costs $0.18 more per gallon than regular, what was the price of each grade of gasoline?
Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0

Hi there! :)

Answer:

Regular = $3.02, Premium = $3.20

------------------------------------------------

Let x = regular gasoline

x + 0.18 = premium gasoline

Write an algebraic equation to express the scenario:

9(x) + 11(x + 0.18) = 62.38

Distribute:

9x + 11x + 11(0.18) = 62.38

Combine like terms and simplify:

20x + 1.98 = 62.38

Subtract 1.98 from both sides:

20x = 60.4

Divide both sides by 20:

x = $3.02

Substitute this value in the equation for premium gasoline:

(3.02) = .18 = $3.20

The prices of each type of gasoline are:

Regular = $3.02, Premium = $3.20

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<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

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X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

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\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

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