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2 years ago

Can someone help me with the equation(s)

Mathematics

1 answer:

Ugo [173]2 years ago

8 0

She purchased 8 sports drinks

{d=p+3

2.25p+1.25d=21.25

d=p+3

p=d - 3

2.25(d - 3) + 1.25d = 21.25

2.25(d-3) + 1.25d=21.25

2.25 - 6.75 + 1.25d = 21.25

3.5d = 21.25+6.75

d=28/3.5

D=8

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If xy + 7ey = 7e, find the value of y'' at the point where x = 0.

Vesnalui [34]

Answer:

Please read my updated answer.

I am wasn't sure what you were asking so here and 2 answers.

1st answer $y=\frac{1}{e}$

2nd answer $\frac{dy}{dx}=-\frac{y}{x+7e}$

Step-by-step explanation:

$xy+7ey=7$

$when \\ x=0$

Substitute $0$ for $x$ then simplify.

$(0*y)+7ey=7\\ 0+7ey=7\\ 7ey=7$

Divide each term in $7ey=7$ by $7e$ and simplify.

$\frac{7ey}{7e}=\frac{7}{7e}$

Cancel the common factor of $7$ on the left side.

Rewrite the expression.

$\frac{ey}{e}=\frac{7}{7e}$

Cancel the common factor of $e$ on the left side.

$y=\frac{7}{7e}$

Cancel the common factor of 7 on the right side.

Rewrite the expression.

$y=\frac{1}{e}$

2nd Answer

Step-by-step explanation:

$xy+7ey=7$

Differentiate both sides of the equation.

$\frac{d}{dx} (xy+7ey)=\frac{d}{dx}(7e)$

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $xy+7ey$ with respect to x is $\frac{d}{dx} [xy]+\frac{d}{dx}[7ey]$

Evaluate $\frac{d}{dx} [xy]$

Differentiate using the Product Rule

Rewrite $\frac{d}{dx} [xy]$ as $y'$ .

$xy'+y+\frac{d}{dx} [7ey]$

Evaluate $\frac{d}{dx} [7ey]$

Since $7e$ is constant with respect to $x$ , the derivative of $7e$ with respect to $x$ is $0$ .

Reform the equation by setting the left side equal to the right side.

$xy'+7ey'+y=0$

Solve for $y'$ .

Subtract $y$ from both sides of the equation.

$xy'+7ey'=-y$

Factor $y'$ out of $xy'$ .

$y'x+7ey'=-y$

Factor $y'$ out of $7ey'$ .

$y'(x)+y'(7e)=-y$

Factor out $y'$ out of $y'(x)+y'(7e)$

$y'(x+7e)=-y$

Divide each term in $y'(x+7e)=-y$ by $x+7e$ and simplify.

$\frac{y'(x+7e)}{x+7e} =\frac{-y}{x+7e}$

Cancel the common factor of $x+7e$ .

$y'=-\frac{-y}{x+7e}$

Replace $y'$ with $\frac{dy}{dx}=-\frac{y}{x+7e}$

$\frac{dy}{dx}=-\frac{y}{x+7e}$

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1 year ago

A submarine first went down 725 feet below sea level. It then went up 150 feet. What was the final depth of the submarine? (A) -

zhannawk [14.2K]

-725+150=-575

therefor the answer is

B)-575

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3 years ago

What is the image of (-2, 3) after a rotation of 90 degrees counterclockwise?

Oduvanchick [21]

Answer:

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Step-by-step explanation:

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2 years ago

15x + 35 = 110 whats the value of x

DerKrebs [107]

answer: 5

hope this helps

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2 years ago

There are two traffic lights on Broadway. The probability of being stopped at the first light is

kiruha [24]

Answer:

Most people found the probability of just stopping at the first light and the probability of just stopping at the second light and added them together. I'm just going to show another valid way to solve this problem. You can solve these kinds of problems whichever way you prefer.

There are three possibilities we need to consider:

Being stopped at both lights

Being stopped at neither light

Being stopped at exactly one light

The sum of the probabilities of all of the events has to be 1 because there is a 100% chance that one of these possibilities has to occur, so the probability of being stopped at exactly one light is 1 minus the probability of being stopped at both lights minus the probability of being stopped at neither.

Because the lights are independent, the probability of being stopped at both lights is just the probability of being stopped at the first light times the probability of being stopped at the second light. (0.4)(0.7) = 0.28

The probability of being stopped at neither is the probability of not being stopped at the first light, which is 1-0.4 or 0.6, times the probability of not being stopped at the second light, which is 1-0.7 or 0.3. (0.6)(0.3) = 0.18

The probability at being stopped at exactly one light is 1-0.18-0.28=.54 or 54%.

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3 years ago