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Wewaii [24]
3 years ago
15

4x-1=2x+11 what is the value of x and what are the steps?

Mathematics
2 answers:
uranmaximum [27]3 years ago
6 0

Answer:

x=6

Step-by-step explanation:

4x-1=2x+11

4x-2x-1=11

2x-1=11

2x=11+1

2x=12

x=12/2

x=6

adelina 88 [10]3 years ago
3 0
X = 6

You need to isolate the variable or get x on one side of the = sign and a number on the other side.

1. Do the inverse operation and add one to each side of the equation. -1+1= 0 and 11+1=12

2. Subtract 2x from both sides. 2x-2x=0 and 4x-2x=2x.

So we are left with 2x=12 or 2 times some number “x” is equal to twelve.

3. Next we divide both sides by 2
2/2=1 and 12/2=6

So x=6
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2 years ago
Using the Segment Addition Postulate, find the value of m.
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36

Step-by-step explanation:

AB + BC = AC

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3 years ago
Plz help!
GuDViN [60]

Answer:

5x - y = 27

Step-by-step explanation:

Slope is the direction of line and it is calculated as,

Slope = \frac{y-y_{1}}{x-x_{1}}


where (x₁, y₁) is any two points on the line.

Here, we have given that

Slope = 5  and (x₁, y₁) = (5, -2)

∴  5 = \frac{y-(-2)}{x-5}


⇒ 5(x - 5) = y + 2

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5 0
3 years ago
(x +y)^5<br> Complete the polynomial operation
Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

\left(x+y\right)^5

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=x,\:\:b=y

=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i

so expanding summation

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

solving

\frac{5!}{0!\left(5-0\right)!}x^5y^0

=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5

=1\cdot \:1\cdot \:x^5

=x^5

also solving

=\frac{5!}{1!\left(5-1\right)!}x^4y

=\frac{5}{1!}x^4y

=\frac{5}{1!}x^4y

=\frac{5x^4y}{1}

=\frac{5x^4y}{1}

=5x^4y

similarly, the result of the remaining terms can be solved such as

\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2

\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3

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=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Therefore,

\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

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