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natita [175]
1 year ago
10

Bonny's truck has a fuel tank which holds 50 gallons, and she typically drives 425 miles on a full tank. She has added a second

tank, which holds 30 gallons. To figure how far she can go between fill-ups now, she multiplies 50 x 425, then divides by 30, to get 708.3 miles. Is this correct?
A) Yes.
B) No; she should have multiplied 425 by 30 + 50.
C) No; she should have divided 425 by 50, then multiplied by 80.
D) No; she should have divided 50 by 30, then multiplied that by 425.
SHOW WORK!!
Mathematics
2 answers:
densk [106]1 year ago
8 0

<u>C</u> is the answer folks.

Usimov [2.4K]1 year ago
6 0
Its C because u have to divide 425 by 50 to get the mpg. Then because the truck now has an 80 gallon tank u have to multiply 8.5 with 80
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4 0
2 years ago
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Lemur [1.5K]

Answer:

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Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

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Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

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Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

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Answer:

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