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gogolik [260]
3 years ago
6

Rewrite the following equation in standard form. -5/6x - 2/5y = 10/3​

Mathematics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

the answer would be 25x+12y= -100

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How do I solve 18=6(2x-8)
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x=11/2

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What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?
bezimeni [28]
3\equiv3\mod{20}
3^2\equiv9\mod{20}
3^3\equiv27\equiv7\mod{20}
3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}

7\equiv7\mod{20}
7^2\equiv49\equiv9\mod{20}
7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}
7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}

9\equiv9\mod{20}
9^2\equiv3^4\equiv1\mod{20}

11\equiv11\mod{20}
11^2\equiv121\equiv1\mod{20}

13\equiv-7\equiv13\mod{20}
13^2\equiv169\equiv9\mod{20}
13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}
13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}

17\equiv-3\equiv17\mod{20}
17^2\equiv(-3)^2\equiv9\mod{20}
17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}
17^4\equiv(-3)^4\equiv81\equiv1\mod{20}

19\equiv-1\equiv19\mod{20}
19^2\equiv19(-1)\equiv-19\equiv1\mod{20}

Generally speaking, a number x coprime to n will be a primitive root of n if we have x^n\equiv x\mod{n}, or x^{n-1}\equiv1\mod{n}. In other words, if x is of order n-1 modulo n, then x is a primitive root of n.

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.
6 0
4 years ago
Need help Now! will brainliest right answer!!
mariarad [96]
A
Hope taht was what you were asking for
3 0
4 years ago
PLEASE HELP ASAP!!!!! WILL GIVE BRAINLEST TO CORRECT ANSWER
marysya [2.9K]
It is B 80. Because it is 40 x 2 = 80. Hope thishelped
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3 years ago
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