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tensa zangetsu [6.8K]
3 years ago
8

3x- 5y=7 5x- 2y=- 1X=? Y=?

Mathematics
1 answer:
Llana [10]3 years ago
3 0
Here is your answer in attachment


hope it help you

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What is the smallest integer k>2000 such that both 17k/66 and 13k/105} are terminating decimals?
Dvinal [7]

17k/66 and 13k/105 must reduce to fractions with a denominator that only consists of powers of 2 or 5.

For example, some fractions with terminating decimals are

1/2 = 0.5

1/4 = 1/2² = 0.25

1/5 = 0.2

1/8 = 1/2³ = 0.125

1/10 = 1/(2•5) = 0.1

1/16 = 1/2⁴ = 0.0625

and so on, while some fractions with non-terminating decimals have denominators that include factors other than 2 or 5, like

1/3 = 0.333…

1/6 = 1/(2•3) = 0.1666…

1/7 = 0.142857…

1/9 = 1/3² = 0.111…

1/11 = 0.09…

1/12 = 1/(2²•3) = 0.8333…

etc.

Since 66 = 2•3•11, we need 17k to have a factorization that eliminates both 3 and 11.

Similarly, since 105 = 3•5•7, we need 13k to eliminate the factors of 3 and 7.

In other words, 17k must be divisible by both 3 and 11, and 13k must be divisible by both 3 and 7. But 13 and 17 are both prime, so it's just k that must be divisible by 3, 7, and 11. These three numbers are relatively prime, so the least positive k that meets the conditions is LCM(2, 7, 11) = 231, and thus k can be any multiple of 231.

If you're familiar with modular arithmetic, this is the same as solving for k such that

13k ≡ 0 (mod 3)

17k ≡ 0 (mod 3)

17k ≡ 0 (mod 7)

13k ≡ 0 (mod 11)

and the Chinese remainder theorem says that k = 231n solves the system of congruences, where n is any integer.

Now it's just a matter of finding the smallest multiple of 231 that's larger than 2000, which easily done by observing

2000 = 8•231 + 152

and so k = 9•231 = 2079.

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Answi do not understand your question

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uranmaximum [27]
x + y = 6 easy medal me plz ? ^^
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Write an expression to represent the sum of three times the square of a number and -7 . In your expression what is the value of
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Answer:

Step-by-step explanation:

3x^2-7

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