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3 years ago

Solve for x. 1/3 x+4=-2/3 x+12

Mathematics

2 answers:

Leona [35]3 years ago

8 0

Answer:

x = 8

Step-by-step explanation:

1. Group all x terms on the left side of the equation

1/3· x+4=-2/3·x+12

Add 2/3x to both sides:

1/3x+4+2/3·x=-2/3x+12+2/3·x

Group like terms:

1/3·x+2/3·x+4=-2/3·x+12+2/3·x

Combine the fractions:

1+2/3·x+4=-2/3·x+12+2/3·x

Combine the numerators:

3/3·x+4=-2/3·x+12+2/3·x

Find the greatest common factor of the numerator and denominator:

1·3/1·3·x+4=-2/3·x+12+2/3·x

Factor out and cancel the greatest common factor:

1x+4=-2/3·x+12+2/3·x

Simplify the left side:

x+4=-2/3·x+12+2/3·x

Group like terms:

x+4=-2/3·x+2/3·x+12

Combine the fractions:

x+4=-2+2/3·x+12

Combine the numerators:

x+4=0/3·x+12

Reduce the zero numerator:

x+4=0x+12

Simplify the arithmetic:

x+4=12

2. Group all constants on the right side of the equation

Subtract 4 from both sides:

x+4-4=12-4

Simplify the arithmetic:

x=12-4

Simplify the arithmetic:

x=8

Mrrafil [7]3 years ago

4 0

Answer:

x=8

Step-by-step explanation:

1/3x +4= -2/3 +12

Subtract 4 from both sides: 1/3x + 4-4 = -2/3x + 12 -4

Simplify: 1/3x = 2/3x + 8

Add 2/3x to both sides: 1/3x + 2/3x = -2/3x + 8 + 2/3x

Simplify: x=8

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anyanavicka [17]

Answer:

24x^2

Step-by-step explanation:

When expanded (x+2)^4= x^4+8x^3+24x^2+32x+16

The only term that matches with these is 24x^2

5 0

3 years ago

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Hi i need help solving this problem and explaing how i got the answer step by step please anyone

Cerrena [4.2K]

Answer:

-161

Step-by-step explanation:

-11(3)-16(8)

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-161

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3 years ago

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4 years ago

I need help in partial fraction!! With simple explaination would be nice!

WITCHER [35]

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}$

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}$

Now the second term can be decomposed into partial fractions.

$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}$
$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}$
$4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)$
$4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1$
$\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3$
$\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}$

So

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}$

3 0

3 years ago

Given the following diagram, if m∠COF = 150°, then m∠BOC =

Naddik [55]

Complete question:
<span>Given the following diagram, if and mCOF = 150°, then mBOC = _____.
150°
90°
45°
30°

I attached a possible diagram that I found in a similar problem.

Using the diagram, I can see that </span>∠COF is supplementary to ∠BOC. This means that the sum of their angles is equal to 180°.

m∠COF + m∠BOC = 180°
m∠COF = 150°

m∠BOC = 180° - 150° = 30° Choice D.

5 0

4 years ago