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borishaifa [10]

3 years ago

7

At a concession stand five hotdogs and two hamburgers cost 8.00 two hotdogs and five hamburgers cost 9.50 find the cost of one h

ot dog and one hamburger

1 answer:

Svetach [21]3 years ago

6 0

Think of this situation as a system of equations.

let y be the price of hotdogs
let x be the price of hamburgers

5y + 2x = 8
2y + 5x = 9.50

now use your solution method of choice
substitution, elimination, matrices, etc

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The amount y of your friend's allowance if the amount she receives is $10 more than the amount x you receive.

velikii [3]

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3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

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

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

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

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

$x^4= (4u)^2$

$x^4= 16u^2$

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

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

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

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

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

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

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

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

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

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Remo needs to buy a countertop for a bathroom. He calculated the area as 10 square feet. The actual area is 9.5 square feet. Wha

Digiron [165]

Answer:

The answer is either 5.26 or 5 I think its 5.26

Step-by-step explanation: The reason why I think its the two answers I said is because 10 is 10 percent of 100 so 9.5 is 9.5 percent which makes me think that

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3 years ago

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Francis have 5 more dollars than Julia together they have at most $45 let X represent the amount of money francis has. Which ine

Dmitrij [34]

X=francis y=julia

the answer is francis has $27.5 and julia has $17.5 you find this by 45 and dividing it by 2 then get on half and add 5 and with the other half subtract five

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3 years ago