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Scilla [17]
3 years ago
5

jess wants to make cupcakes. To make cupcakes, she needs 3 1/4 cups of flour per batch of cupcakes. If Jess has 57 cups of flour

, then how many batches of cupcakes can she make?
Mathematics
2 answers:
inysia [295]3 years ago
7 0
To solve this, we have to divide 57 cups (the total amount she has) by 3.25 (the amount she needs per batch). When we do this, we get 17. 54, but we'll round down because Jess can't make 0.54 of a batch of cupcakes.
Your answer should be 17 batches. I hope this helps!
lawyer [7]3 years ago
4 0
5 is the most amount of batches jess can make with 57 cups of flour
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Ruby plans to make 3 quilts per month until she makes 15 quilts. She figures this will take 1/5 month. Which is NOT a reason why
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Answer: c)She should have multiplied 15 by 3 instead of dividing.


Step-by-step explanation:

Number of quilts Ruby plan to make in one month= 3

which means if Ruby makes more than 3 quilts, the time will be more than 1 month.

To find the number of months to make 15 quilts, she need to divide 15 by 3

Thus, the number of months to make 15 quilts=5

Her answer was 1/5 which was absolutely wrong as if she is making 3 quilt in 1 month then in 1/5  month she would make fewer than 3 quilts.

She divides 3 by 15 instead of 15 by 3, thus but the unit is quilts not months.

Thus the only wrong option is c)She should have multiplied 15 by 3 instead of dividing.


4 0
3 years ago
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Answer:

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Imp gpd 1.2 0.00083 0.00011 1

Step-by-step explanation:

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3 years ago
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gulaghasi [49]

Answer:

The area should be "210 cm to the second power"

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A parabola can be drawn given a focus of (-9, -7) and a directrix of x = 9. Write
Slav-nsk [51]

Check the picture below, so the parabola looks more or less like so, with a vertex at (0 , -7), let's recall the vertex is half-way between the focus point and the directrix.

so this horizontal parabola opens up to the left-hand-side, meaning that the "P" distance is a negative value.

\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

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