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3 years ago

HELP

Mathematics

1 answer:

loris [4]3 years ago

7 0

Answer:

A) s and t.

Step-by-step explanation:

just trust me. Plus lines s and t are the only pair of lines that don't touch each other and can go on forever not touching each other.

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Crazy boy [7]

Answer:

i think the answer is "C" (3-left 2, up 5) i may be wrong sry if i am, i tried :|

Step-by-step explanation:

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2 years ago

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masha68 [24]

Answer:

Step-by-step explanation:

c=37

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4 years ago

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.

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4 years ago

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Serggg [28]

Answer:

Step-by-step explanation:

Since the vertex is 2,1, that's the only equation that works for it. The 2 has to be the opposite, so -2, and the 1 stays 1.

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