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3 years ago

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William got $150 for making all A’s on his report card. He wants to spend it on CDs and DVDs. The local video store is having a

sale; all CDs are $10 and all DVDs are $15. Which of the following is a possible combination of CDs and DVDs?

1 answer:

olga55 [171]3 years ago

3 0

What are the options?

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Help me<br> Help me<br> Help me

koban [17]

Answer:

For the first one is 75 and 30 looks on a calculator add 75+75+30=180

6 0

3 years ago

Sarah Meeham blends coffee for Tasti-Delight. She needs to prepare 120 pounds of blended coffee beans selling for $5.17 per poun

guajiro [1.7K]

Answer:

a) 80 ibs

b) 40 ibs

Step-by-step explanation:

Let X be the pound of the high quality bean at $6.50

(120-x) will be the pound of the cheaper bean at $2.50

6.5x + 2.5(120 - x) = 5.17(120)

6.5x + 300 - 2.5x = 620.4

Collect like terms

6.5x - 2.5x = 620.4 - 300

4x = 320.4

x = 320.4/4

x= 80.1

x = 80 ibs(to the nearest pounds)

For the cheaper bean we have 120 -x

= 120 - 80

= 40 ibs

Sarah would blend 80 ibs of quality bean and 40 ibs of cheaper bean

8 0

3 years ago

Find the value of b in the graph of y=3x+b if it is known that the graph goes through the point: N(0,5)

tino4ka555 [31]

Answer:

I believe b is 5.

Step-by-step explanation:

When you place the points in the values, you get 5=3(0)+b. When you simplify it, you get 5=b. I really hope this is correct, I apologize if it isn't! I hope this helps! :)

4 0

4 years ago

Read 2 more answers

A cracker company packages its product in a box that is shaped like a rectangular prism. The box is 6 inches long, 2 inches wide

Vlada [557]

As the equation is already given, we can just substitute the corresponding values in it!
Volume = 6 x 2 x 8
= 96 in3
So the answer is C. :)

8 0

4 years ago

Read 2 more answers

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

4 years ago