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3 years ago

What is the anwer? plz help. 2(4x+2) = 4x - 12(x-1)

Mathematics

2 answers:

Olin [163]3 years ago

7 0

Answer:

x = 1/2

Step-by-step explanation:

2(4x+2) = 4x - 12(x-1)

Expand brackets.

8x + 4 = 4x - 12x + 12

Subtract 4 on both sides.

8x = 4x - 12x + 12 - 4

Subtract 4x and add 12x on both sides.

8x - 4x + 12x = 12 - 4

Combine like terms.

16x = 8

Divide both sides by 16.

x = 8/16 = 1/2

Juliette [100K]3 years ago

4 0

Answer:

x = 1/2

Step-by-step explanation:

2(4x+2) = 4x - 12(x-1)

Distribute

8x +4 = 4x - 12x +12

Combine like terms

8x + 4 = -8x +12

Add 8x to each side

8x+4+8x = -8x+12 +8x

16x+4 = 12

Subtract 4 from each side

16x +4-4 = 12-4

16x = 8

Divide by 16

16x/16 =8/16

x = 1/2

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Last year, lien earned 1200 more than her husband. Together they earned 45,200. How much did each of them earn?

d1i1m1o1n [39]

Answer:

Lien earned $23,800

Husband earned $22,600

Step-by-step explanation:

$45,200 divide it by 2 = $22,600

then you add $1,200 to $22,600 and get $23,800

4 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$