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Alecsey [184]
2 years ago
14

A gas station sells 1600 gallons of gasoline per hour if it charges $ 2.70 per gallon but only 900 gallons per hour if it charge

s $ 2.95
per gallon. Assuming a linear model
(a) How many gallons would be sold per hour of the price is $ 2.90 per gallon?
Answer:
(b) What must the gasoline price be in order to sell 1300 gallons per hour?
Answer:
Mathematics
1 answer:
kakasveta [241]2 years ago
5 0

Answer:

Look just at finding the linear model. Two points. Use price per gallon for x and gallons per hour sold for y. Your two points (x,y) are (2.15,1600) and (2.55,800).

Slope m=%281600-800%29%2F%280.15-0.55%29

m=800%2F0.40

m=2000

Using point-slope form for a line, and the lower-y point,

y-800=2000%28x-2.55%29

highlight%28y=2000%28x-2.55%29%2B800%29

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(Decimal: 0.07776)

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(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = 12 + 14\; \sqrt{2}.

Step-by-step explanation:

Step One: Simplify the square roots.

8 = 4 \times 2 = 2^2 \times 2.

The square root of 8 \sqrt{8} can be simplified as the product of an integer and the square root of 2:

\sqrt{8} = \sqrt{4 \times 2} = \sqrt{2^2 \times 2} = \sqrt{2^{2}} \times \sqrt{2} = 2 \; \sqrt{2}.

As a result,

(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = (8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2}).

Step Two: Expand the product of the two binomials.

(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2}) is the product of two binomials:

  • Binomial One: 8 - \sqrt{2}.
  • Binomial Two: 2 + 2\; \sqrt{2}

Start by applying the distributive law to the first binomial. Multiply each term in the first binomial (without brackets) with the second binomial (with brackets)

({\bf 8} - {\bf \sqrt{2}}) \cdot {(2 + 2\; \sqrt{2})}\\= [{\bf 8} \cdot {(2 + 2\; \sqrt{2})}] - [{\bf \sqrt{2}} \cdot {(2 + 2\; \sqrt{2})}]

Now, apply the distributive law once again to terms in the second binomial.

[8 \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})] - [\sqrt{2} \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})]\\= [8 \times {\bf 2} + 8 \times {\bf 2\;\sqrt{2}}] - [\sqrt{2} \times {\bf 2} + \sqrt{2} \times {\bf 2\; \sqrt{2}}].

Step Three: Simplify the expression.

The square of a square root is the same as the number under the square root. For example, \sqrt{2} \times \sqrt{2} = (\sqrt{2})^{2} = 2.

[8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + 2 \sqrt{2} \times \sqrt{2}]\\ =[16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4.

Combine the terms with the square root of two and those without the square root of two:

16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2}).

Factor the square root of two out of the second term:

(16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\= (16 - 4) + (16 - 2) \; \sqrt{2} \\= 12 - 14 \; \sqrt{2}.

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(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})\\= [8 \cdot (2 + 2\; \sqrt{2})] - [\sqrt{2} \cdot (2 + 2\; \sqrt{2})]\\= [8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + \sqrt{2} \times 2 \;\sqrt{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times (\sqrt{2})^{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times 2]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\

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