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aliina [53]
2 years ago
6

A man bought a dozen boxes with 24 highlighter pens inside for $8 each box. He repacked five of those boxes into packages of 6 h

ighlighters each and sold them for $3 per package. He sold the rest of the highlighters separately at the rate of 3 pens for $2. How much profit did he make?
Mathematics
2 answers:
guapka [62]2 years ago
5 0
He made 72 dollars of profit
svetoff [14.1K]2 years ago
4 0
He made $72

Hope this helps! :)
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Given the definitions of
Serggg [28]

Answer:

i dont know the question what does it mean

Step-by-step explanation:

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1 year ago
Help please (inequalities)
zaharov [31]

Answer:

The First one is the answer

4 0
2 years ago
Read 2 more answers
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
What is the midpoint of a segment with endpoints at(-4,-8) and (8, 10)?
vekshin1

Answer:

The answer is

<h2>( 2 , 1)</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

M = ( \frac{x1 + x2}{2}  , \:  \frac{y1 + y2}{2} )

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4,-8) and (8, 10)

The midpoint is

M = ( \frac{ - 4 + 8}{2}  , \:  \frac{ - 8 + 10}{2} ) \\  = ( \frac{4}{2} ,  \frac{2}{2} )

We have the final answer as

<h3>( 2 , 1)</h3>

Hope this helps you

4 0
3 years ago
35=11x. i need help
V125BC [204]

Answer:

x= 35/11

Step-by-step explanation:

steps

35 = 11x

swich sides

11x = 35

Divide both sides by 11

11x/11 = 35/11

simplify

x = 35/11

5 0
2 years ago
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