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3 years ago

3+2= 2x=6 Please help me

2 answers:

5 0

Well.
3+2 = 5
And 2x = 6 is a simple equation.

Our main goal is to isolate the x (get the x to be alone)
Our first step is to divide both sides by 2
2x = 6
—- —-
2 2

Both 2’s on the left side cancel out. (Disappear)
And we solve on the right side.
6/2 = 3
Answer: x = 3

GOOD LUCK!

oksano4ka [1.4K]3 years ago

4 0

The first one is 5 and the second is x=3

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m_a_m_a [10]

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6 0

3 years ago

What is the slope of the line?

Natasha_Volkova [10]

y2 - y1 / x2 - x1 = slope

2 - -2 / 2 - -1

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8 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

Solve for the equation for x 3x+y/=2

avanturin [10]

Answer: x= 2/3 - 1/3y

Explanation step by step:

Step 1: Move variable to the right hand side and change its sign.

3x = 2-y

Step 2: divide both sides of the equation by 3.

x= 2/3 - 1/3y

3 0

3 years ago

PLS HELP, HAVE A GOOD DAY

Snowcat [4.5K]

Answer:

$\frac{\pi }{4} +\frac{\pi }{2} n,$ where n∈Z.

Step-by-step explanation:

2sin²x-1; ⇒ sin²x=1/2;

$\left[\begin{array}{ccc}sinx=-\frac{1}{\sqrt{2}} \\sinx=\frac{1}{\sqrt{2}} \end{array} \ = > \ \left[\begin{array}{ccc}x=(-1)^{l+1}\frac{\pi }{4}+\pi l \\x=(-1)^k\frac{\pi }{4}+\pi k \end{array} \ = > x=\frac{\pi }{4} +\frac{\pi }{2}n, \ n-Z.$

4 0

2 years ago

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