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taurus [48]
3 years ago
11

A recipe requires 1/3 cup of milk for each 1/4 cup of water. How many cups of water are needed for each cup of milk?

Mathematics
2 answers:
goldfiish [28.3K]3 years ago
8 0
D) 1 1/3

(1/3) x reciprocal of 1/4 so (4/1) = 4/3 which simplifies to 1 1/3 
kenny6666 [7]3 years ago
6 0
\bf \begin{array}{ccllll}
milk&water\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
\frac{1}{3}&\frac{1}{4}\\
1&x
\end{array}\implies \cfrac{\frac{1}{3}}{1}=\cfrac{\frac{1}{4}}{x}\implies \cfrac{\frac{1}{3}}{\frac{1}{1}}=\cfrac{\frac{1}{4}}{\frac{x}{1}}
\\\\\\
\cfrac{1}{3}\cdot \cfrac{1}{1}=\cfrac{1}{4}\cdot \cfrac{1}{x}\implies \cfrac{1}{3}=\cfrac{1}{4x}\implies 4x=3\implies x=\cfrac{3}{4}
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Answer:

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Step-by-step explanation:

The general term of an arithmetic sequence is ...

  an = a1 +d(n -1)

Then the ratio of interest is ...

  (7th term)/(9th term) = 5/8

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Multiplying by 8(a1 +8d) we get ...

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The common difference is -3/8 times the first term.

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<u>Example</u>

Let the first term be -16. Then the common difference is (-3/8)(-16) = 6. The first 9 terms of the sequence are ...

  -16, -10, -4, 2, 8, 14, 20, 26, 32

The ratio of the 7th and 9th terms is ...

  20/32 = 5/8 . . . . as required

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In the general case, the ratio of terms would be ...

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