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3 years ago

Eleven more than twice x is 5 less than x.

Mathematics

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

x= -2

Step-by-step explanation:

11+2x = 5 - x

6 = -3x

x = -2

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4x + 10 = 2x - 14<br> Find the value of x.

Gre4nikov [31]

Answer:

x= -12

Step-by-step explanation:

Simplifying

4x + 10 = 2x + -14

Reorder the terms:

10 + 4x = 2x + -14

Reorder the terms:

10 + 4x = -14 + 2x

Solving

10 + 4x = -14 + 2x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-2x' to each side of the equation.

10 + 4x + -2x = -14 + 2x + -2x

Combine like terms: 4x + -2x = 2x

10 + 2x = -14 + 2x + -2x

Combine like terms: 2x + -2x = 0

10 + 2x = -14 + 0

10 + 2x = -14

Add '-10' to each side of the equation.

10 + -10 + 2x = -14 + -10

Combine like terms: 10 + -10 = 0

0 + 2x = -14 + -10

2x = -14 + -10

Combine like terms: -14 + -10 = -24

2x = -24

Divide each side by '2'.

x = -12

Simplifying

x = -12

3 0

3 years ago

Read 2 more answers

Below are two parallel lines intersecting them

slavikrds [6]

Answer:

Step-by-step explanation:

corresponding angle to 56 is also opposite to x which means it is also 56

3 0

3 years ago

Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

6 0

4 years ago

Plz no link i need real answers please

leonid [27]

Answer:

You can download the link here:

https://rb.gy/tmu0gl

JKJKJKJKJKJKJKJKJKJKJK

Your answer is n = 9/8

Brainliest? Thx.

3 0

3 years ago

How do i find the congruence of triangles??? i dont know how to do it at all please help

Elza [17]

Step-by-step explanation:

"If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent. If any two angles and the included side are the same in both triangles, then the triangles are congruent."

6 0

3 years ago