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gulaghasi [49]
3 years ago
8

How many solutions does the equation have? 6-a = 1/3(9a-18)

Mathematics
1 answer:
Alex777 [14]3 years ago
5 0
6 - a = \frac{1}{3} (9a - 18)

First, simplify \frac{1}{3} (9a - 18) to \frac{9a-18}{3} / Your problem should look like: 6 - a =  \frac{9a - 18}{3}
Second, multiply both sides by 3. / Your problem should look like: 18 - 3a = 9(a - 2)
Third, expand. / Your problem should look like: 18 - 3a = 9a - 18
Fourth, add 3a to both sides. / Your problem should look like: 18 = 9a - 18 + 3a
Fifth, simplify 9a - 18 + 3a to 12a - 18. / Your problem should look like: 18 = 12a - 18
Sixth, add 18 to both sides. / Your problem should look like: 18 + 18 = 12a
Seventh, add 18 + 18 to get 36. / Your problem should look like: 36 = 12a
Eighth, divide both sides by 12. / Your problem should look like: \frac{36}{12} = a
Ninth, since 3 goes into 12 to get 36, simplify \frac{36}{12} to 3. / Your problem should look like: 3 = a
Tenth, switch sides. / Your problem should look like: a = 3

Answer: a = 3 (one solution)

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Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

\displaystyle \int x^3 \ln(x)^2 \, dx

Integrate by parts:

\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du

where

u = ln(x)²   ⇒   du = 2 ln(x)/x dx

dv = x³ dx   ⇒   v = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx

Integrate by parts again:

\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'

where

u' = ln(x)   ⇒   du' = dx/x

dv' = x³ dx   ⇒   v' = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx

So, we have

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C

\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}

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What is the value of 4(3x + 5) when x = 11?
navik [9.2K]

Answer:

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Step-by-step explanation:

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4(33+5)

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3 years ago
Write 1/4 as a decimal.
n200080 [17]

Answer:

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Step-by-step explanation:

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If you want to simplify, 2/4 divided by 2 equals 1/2.

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Simplified: 1/2

Hope I was able to help!
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