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mrs_skeptik [129]
4 years ago
8

Joaquin has created a small treasure map that is in the shape of a square. He decides to make a larger version of the map, and o

n this new map he increases each side by 6 inches. The area of this larger version of the map is 121 square inches. What are the dimensions of his original, smaller map? 5 inches by 5 inches 8 inches by 8 inches 11 inches by 11 inches 17 inches by 17 inches
Mathematics
2 answers:
likoan [24]4 years ago
9 0
A - side of small map
a+6 - side of larger map
(a+6)² - area of larger map

(a+6)² = 121
(a+6)² = 11²    |√(...)
a+6 = 11    |-6
a+6-6 = 11-6
<span>a = 5

</span>Answer A. <span>5 inches by 5 inches</span>
Furkat [3]4 years ago
6 0

Answer:

A 5 inches by 5 inches

Step-by-step explanation:

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Given, expression is 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)

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Now, take the given expression.

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Rearranging the terms we get,

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\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)

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\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.

\text { We know that } \log a-\log b=\log \frac{a}{b}

\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)

Hence, the simplified form \rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)

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