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Alex73 [517]
2 years ago
6

Eliza's backpack weighs 18 and StartFraction 7 over 9 EndFraction pounds with her math book in it. Without her math book, her ba

ckpack weighs 14 and StartFraction 7 over 8 EndFraction pounds. How much does Eliza's math book weigh? 3 and StartFraction 11 over 72 EndFraction pounds 3 and StartFraction 65 over 72 EndFraction pounds 4 and StartFraction 11 over 72 EndFraction pounds 4 and StartFraction 65 over 72 EndFraction pounds
Mathematics
1 answer:
kati45 [8]2 years ago
7 0

Answer:

Math\ Book = 3\frac{65}{72}\ kg

Step-by-step explanation:

Given

<em />Backpack = 18\frac{7}{9}kg<em> --- with math book</em>

<em />Backpack = 14\frac{7}{8}kg<em> --- without math book</em>

<em />

Required

Determine the weight of her math book

To do this, we simply calculate the difference in the weights given.

Math Book= Backpack(with math book) - Backpack (without math book)

Math\ Book = 18\frac{7}{9} - 14\frac{7}{8}

Math\ Book = \frac{169}{9} - \frac{119}{8}

Take LCM

Math\ Book = \frac{8 * 169 - 9 * 119}{72}

Math\ Book = \frac{1352 - 1071}{72}

Math\ Book = \frac{281}{72}

Math\ Book = 3\frac{65}{72}\ kg

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For k=5

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

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For k=6

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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