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ehidna [41]
3 years ago
14

Write 81.42 as a mixed number

Mathematics
2 answers:
Kruka [31]3 years ago
7 0
81.42 =81 \frac{42}{100}=81\frac{21\times 2}{50\times 2} = \boxed{\bf{81\frac{21}{50}}}
Effectus [21]3 years ago
5 0
81.42
= 81+ 0.42
= 81+ 42/100
= 81+ (42/2) / (100/2)
= 81+ 21/50
= 81 21/50

The final answer is 81 21/50~
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3 years ago
WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"
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Recall Euler's theorem: if \gcd(a,n) = 1, then

a^{\phi(n)} \equiv 1 \pmod n

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We have \gcd(9,32) = 1 - in fact, \gcd(9,32^k)=1 for any k\in\Bbb N since 9=3^2 and 32=2^5 share no common divisors - as well as \phi(9) = 6.

Now,

37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} =  32\cdot\left(32^{(\cdots)}\right)^6

where the c_i are positive integer coefficients from the binomial expansion. By Euler's theorem,

\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9

so that

32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9

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2 years ago
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I am Lyosha [343]

Answer:

70/29 or 2 12/29

Step-by-step explanation:

5 0
2 years ago
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