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4 years ago

Firat cheese yeur beunda and intervals by completing the

Mathematics

1 answer:

katovenus [111]4 years ago

7 0

Wnwnejeidjejebehsjdeijebebebsjxidjebebe

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Please hurry find the measure of P

Inessa [10]

Answer:

m∠P = 65°

Step-by-step explanation:

An Isosceles triangle has two sides of equal length and two interior angles of equal measure.

Sum of interior angles of a triangle = 180°

If QO ≅ PQ and the enclosed m∠Q = 50° then m∠P ≅ m∠O

⇒ m∠Q + m∠P + m∠O = 180

⇒ 50 + m∠P + m∠O = 180

⇒ m∠P + m∠O = 130

As m∠P ≅ m∠O:

⇒ m∠P = 130 ÷ 2 = 65°

7 0

2 years ago

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2. Joseph scuba dives to a depth of 35 feet below sea level. He then ascends

Alex777 [14]

If he is already 35 feet below sea level, then he goes across 10 feet. Take 35+22, which gives you 57. Joseph’s current depth is 57 feet down.

3 0

3 years ago

An Xbox is $399.79, if

Llana [10]

A. $19.98
b. $419.77
i think thats the answer

7 0

3 years ago

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I need help on number 5 for my practice exam

kenny6666 [7]

Answer:

C is the correct answer

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3 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.

6 0

4 years ago