Mathematics: The owner of a 5000-gal oil truck loads the truck with gasoline and kerosene. The profit on each gallon of gasoline is 16¢ and on each gallon of kerosene it is 15¢. How many gallons of each fuel did the owner load if the profit was $780?

The owner of a 5000-gal oil truck loads the truck

Ответ:

devilrao6742

28.01.2022

3000 gallons of gasoline and 2000 gallons of kerosene

Step-by-step explanation:

The first thing is that 16 ¢ and 15 ¢ are 16 and 15 cents, therefore we can express it as $ 0.16 and $ 0.15 respectively.

Now, we can solve by means of a 2x2 system of equations, we have to:

"x" is the number of gallons of gasoline

"y" be the number of gallons of kerosene

x + y = 5000 => x = 5000 - y

0.16 * x + 0.15 * y = 780

replacing

0.16 * (5000 - y) + 0.15 * y = 780

800 - 0.16 * y + 0.15 * y = 780

-0.01 * y = 780 - 800

y = -20 / -0.01

y = 2000

Therefore x:

x = 5000 - 2000

x = 3000

which means 3000 gallons of gasoline and 2000 gallons of kerosene

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Ответ:

viridianasar5158

22.10.2020

Mathematics

So the answer is DNC(Does Not Convergence)

Step-by-step explanation:

Given;

$\int\limits^\infty_5 {\frac{7}{x^{2}-9 } } \ dx$

We know,

The integral is an improper integral because the upper limit of integration approaches infinity. First, replace the infinite upper limit by the finite limit 'l' and take the limit of 'l' to approach infinity.

$\int\limits^\infty_5 {\frac{7}{x^{2}-9 } } \, dx=7 \lim_{l \to \infty} \int\limits^\infty_5 {\frac{1}{x^{2}-9 } } \, dx$

$7 \lim_{l \to \infty} \int\limits^\infty_5 {\frac{1}{x^{2}-3^{2} } } \, dx$

$7\times \lim_{l \to \infty} \left [ \frac{1}{2\times 3} \log\left |\frac{x-3}{x+3} \right | \right ]_{5}^{l}$ (∵We know; $\int {\frac{1}{x^{2}-a^{2} } } \, dx= \frac{1}{2\times a} \log\left |\frac{x-3}{x+3} \right |$ )

$\frac{7}{6} \times \lim_{l \to \infty} \left[ \log\left | \frac{l-3}{l+3} \right |-\log\left | \frac{5-3}{5+3}\right |\right]$

$\frac{7}{6} \times \lim_{l \to \infty} \left[ \log\left | \frac{l-3}{l+3} \right |-\log\left | \frac{-2}{8}\right |\right]$

$\frac{7}{6} \times \left[ \log\left | \frac{\infty-3}{\infty+3} \right |-\log( \frac{2}{8})\right]$ (∵ $\left | -\frac{2}{8} \right |=(\frac{2}{8} )$ )

$\frac{7}{6} \times \left[ \infty-\log( \frac{2}{8})\right]$

$\frac{7}{6}\times \infty$

$\infty$

Thus the integral Does Not Convergence.

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The owner of a 5000-gal oil truck loads the truck with gasoline and kerosene. The profit on each gallon of gasoline is 16¢ and on each gallon of kerosene it is 15¢. How many gallons of each fuel did the owner load if the profit was $780?

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