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3 years ago

Someone help please?

Mathematics

2 answers:

iren [92.7K]3 years ago

8 0

It get you too x-3 will give you 9 f(x) -9-3 gives you 43:76

Yuliya22 [10]3 years ago

6 0

Answer:

43:76

Step-by-step explanation:

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Help plz! Please explain how u solved it i need as much help as possible

Stells [14]

What do you need help with?

7 0

3 years ago

An unknown number y is 20 more than an unknown number x. The number y is also x less than 2. The equations to find x and y are s

KatRina [158]

The graph is given below. Then the intersection point will be (-9, 11).

<h3>What is the linear system?</h3>

A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.

An unknown number y is 20 more than an unknown number x.

The number y is also x less than 2.

The equations to find x and y are shown below:

y = x + 20 ...1

y = −x + 2 ...2

Add both equations, then we have

2y = 22

y = 11

Then put the value of y in equation 1, then we get

11 = x + 20

x = 11 - 20

x = -9

Then the intersection point will be (-9, 11)

More about the linear system link is given below.

brainly.com/question/20379472

#SPJ1

4 0

2 years ago

According to the diagram, which are congruent? check all that apply.

Andre45 [30]

Sorry, I can’t help you. You forgot to include a photo of the diagram.

5 0

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

How many decimals are between 0,5 and 0,6 ?

dusya [7]

Between 0.5 and 0.6 can be an infinity.

<em><u>ex:</u></em><u /><u><em /></u><em />

0.51

0.52

0.517

0.5169889549.......

0.5999999999..........

etc.

5 0

3 years ago

Read 2 more answers