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Svetllana [295]
3 years ago
9

Lupe bought 8 pounds of pretzels at a local wholesaler for $18. Her friend Charles bought 6 pounds of pretzels at the supermarke

t for $15. Charles thinks he got the better deal because $15 is less than $18
How much would Charles have spent at the supermarket if he had purchased the same amount of pretzels as Lupe bought at the wholesaler? Show your work explain your answer, in a paragraph
Mathematics
2 answers:
Novay_Z [31]3 years ago
3 0

Answer:

20

Step-by-step explanation:

Charle would have spent that much because one pound is 2.50

15 divided by 6=2.50

2.50x8=20

alexdok [17]3 years ago
3 0

Answer:

Charles would've spent $20 at the supermarket if he bought the same amount of pounds as Lupe.

Step-by-step explanation:

18÷8= 2.25

2.25 is the amount of money per pound at the wholesaler.

15÷6= 2.50

2.50 is the amount of money per pound at the supermarket.

2.50x8=  20

Charles would've spent $20 if he bought the same amount of pounds as Lupe. Since one pound cost 2.50 at the supermarket, 8 times 2.50 is 20. He would actually spend more at the supermarket than the wholesaler.

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What is the value of x <br> Enter your answer, as a decimal in the box
Maru [420]

Answer:

the answer is 8.334

Step-by-step explanation:

3 0
3 years ago
) 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
A chemist needs to create a 50% HCl solution. (HCl is hydrochloric acid. A "50% HCl solution" contains 50% HCl and the other 50%
lorasvet [3.4K]

25/100=30/x x=120 (30+70y)/(120+100y)=50/100 100y=?

7 0
3 years ago
2.
Lynna [10]

In 15 minutes, Jada would have covered about 2.5 miles, while the cousin would have covered 0.625 miles.

<h3>What is speed?</h3>

Speed is the ratio of distance to time.

For Jada:

Speed = 2 miles / 12 minutes = 0.167 miles per minute

For Jada cousin:

Speed = 1 mile / 24 min = 0.0417 miles per minute.

Hence, in 15 minutes:

Jada distance = 15 min * 0.167 miles per minute = 2.5 miles

Jada cousin distance = 15 min * 0.167 miles per minute = 0.625 miles

In 15 minutes, Jada would have covered about 2.5 miles, while the cousin would have covered 0.625 miles.

Find out more on speed at: brainly.com/question/4931057

6 0
2 years ago
ALGEBRA 1, SEMESTER 2
SIZIF [17.4K]

So firstly, we have to find f(x) when x = 8 and x = 0. Plug the two numbers into the x variable of the function to solve for their f(x):

f(0)=\frac{2*0+3}{3*0-3}\\ f(0)=\frac{0+3}{0-3}\\ f(0)=\frac{3}{-3}\\ f(0)=-1\\ \\ f(8)=\frac{2*8+3}{3*8-3}\\ f(8)=\frac{16+3}{24-3}\\ f(8)=\frac{19}{21}

Now that we have their y's, we can use the slope, aka average rate of change, formula, which is \frac{y_2-y_1}{x_2-x_1} . Using what we have, we can solve it as such:

\frac{\frac{19}{21}-(-1)}{8-0}=\frac{\frac{40}{21}}{8}=\frac{40}{21*8}=\frac{40}{168}=\frac{5}{21}

In short, the average rate of change from x = 0 to x = 8 is 5/21.

8 0
3 years ago
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