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RUDIKE [14]
3 years ago
15

Natalie has 300 cookies for a bake sale. She makes 8 more batches with 12 cookies in each batch. She sells 6 bags of cookies wit

h 21 cookies in each bag. Natalie says she has about 100 cookies left to sell.
Is Natalie’s estimate reasonable?


Natalie’s estimate is not reasonable. She makes about 100 more cookies and sells about 100 cookies so she should have about 300 left.

Natalie’s estimate is not reasonable. She makes about 100 more cookies and sells about 100 cookies so she should have about 100 fewer cookies than she started with.

Natalie’s estimate is reasonable. She makes about 200 more cookies and sells about 100 cookies so she should have about 100 left.

Natalie’s estimate is not reasonable. She makes about 200 more cookies and sells about 100 cookies so she should have about 100 more than she started with.
Mathematics
1 answer:
vitfil [10]3 years ago
6 0

Answer:

I would say it's not reasonable. the answer would be the first one

Step-by-step explanation:

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Let width = x

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\begin{aligned} \implies \textsf{Area of original picture} & = \sf width \times length\\& = x\left(\dfrac{3}{5}x\right)\\& = \dfrac{3}{5}x^2\end{aligned}

We are told the area of the enlarged picture is 375 in².  Therefore, substitute this into the equation and solve for x to find the width of the enlarged picture:

\begin{aligned}\textsf{Area} & = 375\\ \implies \dfrac{3}{5}x^2 & = 375\\ x^2 & =375 \cdot \dfrac{5}{3}\\ x^2 & =625\\ x & = \sqrt{625}\\ x& = 25\end{aligned}

Therefore, the width of the enlarged picture is 25 in.

Substitute the found value of x into the expression for length to find the length of the enlarged picture:

\begin{aligned}\sf Length & = \dfrac{3}{5}x\\\\\implies \sf Length & = \dfrac{3}{5}(25)\\\\& = 15\: \sf in\end{aligned}

Therefore, the dimensions of the enlarged picture are <u>25 in x 15 in</u>.  The width is 25 in and the length is 15 in, as the length is 3/5 of the width.

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