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creativ13 [48]
3 years ago
9

What is the factorization of 729^15+1000​

Mathematics
1 answer:
Nesterboy [21]3 years ago
3 0

\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 729=27^2\\ \qquad (3^3)^2\\ 1000=10^3 \end{cases}\implies 729^{15}+1000\implies ((3^3)^2)^{15}+10^3 \\\\\\ ((3^2)^{15})^3+10^3\implies (3^{30})^3+10^3\implies (3^{30}+10)~~[(3^{30})^2-(3^{30})(10)+10^2] \\\\\\ (3^{30})^3+10^3\implies (3^{30}+10)~~~~[(3^{60})-(3^{30})(10)+10^2]

now, we could expand them, but there's no need, since it's just factoring.

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The answer is,

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3 years ago
4, Find a number x such that x = 1 mod 4, x 2 mod 7, and x 5 mod 9.
olchik [2.2K]

4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.

Start with

x=7\cdot9+4\cdot2\cdot9+4\cdot7\cdot5

Taken mod 4, the last two terms vanish and we're left with

x\equiv63\equiv64-1\equiv-1\equiv3\pmod4

We have 3^2\equiv9\equiv1\pmod4, so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.

x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5

Taken mod 7, the first and last terms vanish and we're left with

x\equiv72\equiv2\pmod7

which is what we want, so no adjustments needed here.

x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5

Taken mod 9, the first two terms vanish and we're left with

x\equiv140\equiv5\pmod9

so we don't need to make any adjustments here, and we end up with x=401.

By the Chinese remainder theorem, we find that any x such that

x\equiv401\pmod{4\cdot7\cdot9}\implies x\equiv149\pmod{252}

is a solution to this system, i.e. x=149+252n for any integer n, the smallest and positive of which is 149.

3 0
3 years ago
John left school with $8.43. He found a quarter on his way home and then stopped to buy an apple for $0.89. How much money did h
Marina86 [1]

Answer:

He had $7.79.

Step-by-step explanation:

8.43 + 0.25 = 8.68

8.68 - 0.89 = 7.79

5 0
3 years ago
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Delicious77 [7]
B is the correct answer



\frac{x}{100}  =  \frac{53}{106}  \\ x =  \frac{53 \times 100}{106}  =  \frac{5300}{106}  = 50




good luck
4 0
3 years ago
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V125BC [204]
I thank it is B you have to add all the tickets and times that by the the amount of money  
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3 years ago
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