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2 years ago

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On Sunday, a local hamburger shop sold a combined of 572 hamburger and cheeseburger. The number of cheeseburgers sold was three

times the number of hamburger sold. How many hamburger were sold on Sunday

1 answer:

igor_vitrenko [27]2 years ago

3 0

Answer: 143 hamburgers and 429 cheese burgers

Explanation:

Call h and c the number of both items.
(h-hamburger and c-cheeseburger)

h + c = 572
c = 3h

Sub the second into the first

h + 3h = 572
4h = 572

Divide both sides by 4
h = 143 hamburgers

Use this back into the second equation
c = 3 • 143 = 429 cheeseburgers

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Answer:

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Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

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Differentiate

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$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

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Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

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Simplify

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In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

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$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

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Andreyy89

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