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BARSIC [14]
3 years ago
6

How to do 2x+4=12 plz show how you got it

Mathematics
2 answers:
OverLord2011 [107]3 years ago
8 0
2x + 4 = 12

We're simply just trying to isolate x.

So, we must get x onto it's own side of the equal sign :)

Our first step is to subtract 4 from both sides.

2x + 4 - 4 = 12 - 4

Simplify.

2x = 8

Then, we divide both sides by 2.

2x ÷ 2 = 8 ÷ 2 

Simplify.

x = 4

----------

To check your work, simply plug in the value of x into x in the original equation.

In this problem, x = 4, so plug in 4 for x.

2x + 4 = 12

2(4) + 4 = 12

Simplify.

8 + 4 = 12

12 = 12

Therefore, x = 4

~Hope I helped!~
BlackZzzverrR [31]3 years ago
6 0
First you subtract 4 from both sides.
2x+4-4=12-4
Simplify
2x=8
Divide both sides by 2 
\frac{2x}{2} and \frac{8}{2}

Final answer {x=4}
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harkovskaia [24]

Answer:

15x^2-x-2

Step-by-step explanation:

that's you're answer

5 0
3 years ago
An angle, θ, is in standard position. The terminal side of the angle passes through the point (4, 5).
lara31 [8.8K]

Answer:

cos Θ = \frac{4}{\sqrt{41} }

Step-by-step explanation:

Given the angle passes through (4, 5 ) then the right triangle with sides 4 and 5 has hypotenuse h given by

h² = 4² + 5² = 16 + 25 = 41 ( take the square root of both sides )

h = \sqrt{41}

In relation to Θ, 4 is adjacent and 5 is opposite, thus

cos Θ = \frac{adjacent}{hypotenuse} = \frac{4}{\sqrt{41} }

5 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
9<br><img src="https://tex.z-dn.net/?f=9x%20%2B%201%20%20%2B%2032x%20%2B%201%20%3D%2036%20%5C%5C%20" id="TexFormula1" title="9x
AysviL [449]
9x+1+32x+1=36

Combine like terms

41x+2=36

Subtract 2 from both sides

41x=34

X= 41/34 or 1.2
7 0
2 years ago
My last 2 questions so I felt generous. (Answer them for brainiest)
prohojiy [21]

Answer:

Step-by-step explanation:

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20. 45000+300m=60000

m=50

45000+300m>60000  

He needs at least 50 machines to make the 60000.

5 0
3 years ago
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