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3 years ago

6

A store charges a restocking fee for any returned item based upon the item price. An item priced at $200 has a fee of $12. An it

em priced at $150 has a fee of $9. What percentage of the item price is the restocking fee?

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

6%

Step-by-step explanation:

for this is divided the fee by the price of the item.

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Which of the following values are in the range of the function graph below check all that apply

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It is
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3 years ago

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Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin

Anastasy [175]

Suppose $y_2(x)=y_1(x)v(x)$ is another solution. Then

$\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}$

Substituting these derivatives into the ODE gives

$x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0$

$x^5v''+5x^4v'=0$

Let $u(x)=v'(x)$ , so that

$\begin{cases}u=v'\\u'=v''\end{cases}$

Then the ODE becomes

$x^5u'+5x^4u=0$

and we can condense the left hand side as a derivative of a product,

$\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C$

$x^5u=C\implies u=Cx^{-5}$

Solve for $v$ :

$v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2$

Solve for $y_2$ :

$\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}$

So another linearly independent solution is $y_2=\dfrac1x$ .

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3 years ago

Please help il give u anything !!

makvit [3.9K]

Answer:

Step-by-step explanation:

Woah thats hard what year you in?

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3 years ago

What is an equation of the line that passes through the points (-6, -2) and

den301095 [7]

Answer:

The equation of line is: $\mathbf{4x-3y=-18}$

Step-by-step explanation:

We need to find an equation of the line that passes through the points (-6, -2) and (-3, 2)?

The equation of line in slope-intercept form is: $y=mx+b$

where m is slope and b is y-intercept.

We need to find slope and y-intercept.

Finding Slope

Slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have $x_1=-6,y_1=-2, x_2=-3, y_2=2$

Putting values and finding slope

$Slope=\frac{2-(-2)}{-3-(-6)}\\Slope=\frac{2+2}{-3+6} \\Slope=\frac{4}{3}$

So, we get slope: $m=\frac{4}{3}$

Finding y-intercept

Using point (-6,-2) and slope $m=\frac{4}{3}$ we can find y-intercept

$y=mx+b\\-2=\frac{4}{3}(-6)+b\\-2=4(-2)+b\\-2=-8+b\\b=-2+8\\b=6$

So, we get y-intercept b= 6

Equation of required line

The equation of required line having slope $m=\frac{4}{3}$ and y-intercept b = 6 is

$y=mx+b\\y=\frac{4}{3}x+6$

Now transforming in fully reduced form:

$y=\frac{4x+6*3}{3} \\y=\frac{4x+18}{3} \\3y=4x+18\\4x-3y=-18$

So, the equation of line is: $\mathbf{4x-3y=-18}$

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2 years ago