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Bond [772]
3 years ago
12

A store has apples on sale for $3.00 for 2 pounds. How many pounds of apples can you buy for $9? If an apple is approximately 5

ounces, how many apples can you buy for $9? Explain your reasoning.
Mathematics
2 answers:
vazorg [7]3 years ago
5 0
Since 2 pounds is $3 and you have $9, you would do: 

\frac{3}{2} =  \frac{9}{x} , x= # of pounds of apples

You do cross multiplication to get:

3x=18

x=6      

1 lb= 16 oz

So 6*16=96   to get the number of ounces you can buy for $9.

Then divide 96 by 5, so that is ABOUT 19 APPLES.


kobusy [5.1K]3 years ago
4 0

we know that

A store has apples on sale for $3.00 for 2 pounds

Part a) How many pounds of apples can you buy for $9?

by using proportion

\frac{3}{2}\frac{usd}{pound}  =\frac{9}{x} \frac{usd}{pound} \\ \\ 3x=9*2\\ \\ x=\frac{18}{3} \\ \\ x=6 pounds

therefore

the answer Part a) is

6 pounds

Part b) If an apple is approximately 5 ounces, how many apples can you buy for $9?

we know that

1 pound=16 ounces

step 1

convert ounces to pounds

1apple=\frac{5}{16}pounds

by using proportion

\frac{1}{\frac{5}{16}}\ \frac{apples}{pounds} =\frac{x}{6}  \frac{apples}{pounds}\\ \\ x\frac{5}{16} =6\\ \\ x=\frac{16*6}{5} \\ \\ x=19.2 apples

therefore

the answer part b) is

about 19 apples

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At what point does she lose contact with the snowball and fly off at a tangent? That is
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Answer:

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight  down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the  instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to  the skier make with the vertical?"

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- If we ignore snow-­ski friction along with air resistance, then the one work producing force in this problem, gravity,  is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical  energy at the top of the snowball.

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- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity  can hold her to that path, i.e. as long as:

                         m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal  force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

                            3*cos(α) ≥ 2

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