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3 years ago

The job paid $25 for every 2 hours of work. Write an equation that represents how much the jobs pays, y, for x hours of work.

Mathematics

1 answer:

sergejj [24]3 years ago

6 0

Answer:

$12,5xy

Step-by-step explanation:

y = how much the jobs pays

x = hours of work

$25 \div 2$

$= 12.5xy$

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What is 12+12 and give good reson why

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

you have 12 of something and Are adding 12 to it.

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3 years ago

What is 15 divided by 105

shusha [124]

Answer:

Step-by-step explanation:

you can figure this out by taking 105 and keep subtracting 15 until you reach 0 then see how many times you needed to use 15

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3 years ago

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If r and s are positive integers, each greater than 1, and if 11(s-1) =13(r-1), what is the least possible value of (r+s)?

Tresset [83]

Answer:

The least value of (r+s) is (12+14)=26

Step-by-step explanation:

Well, first let us solve equation of 11(s-1)=13(r-1), which results in 11s-11=13r-13. Hence, 11s+2=13r. It is stated that r and s both are integers and greater than 1.

To make sure that r and s are integers, the least value of s must be equal to 14 (s=14) then the least value of r becomes 12 (r=12).

Finally, the least value of (r+s) is (12+14)=26.

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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

mr. duc asked perry to set up 450 chairs in a large hall. perry puts 20 chairs in each row. he has already set up 15 rows of cha

Brums [2.3K]

The number of chairs that still need to be set up are 150.

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4 years ago

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