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3 years ago

What is the difference between 6z and z6

Mathematics

2 answers:

I am Lyosha [343]3 years ago

4 0

Since multiplication is dissolved in both monomials, there is no difference.

ValentinkaMS [17]3 years ago

3 0

6z - z^6

Since we don't have the value for z, we cannot solve the expression, however, we can get out most simple expression.

Answer: 5z^6

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Integrala x la a treia ori ln la a doua dx va rog

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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2 years ago

someone came into my shop and stole $100.00 from the register without my knowledge. The person came back in with the same $100.0

Vaselesa [24]

100 bucks hope this helps

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3 years ago

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Estimate the product by rounding 9 times 54

crimeas [40]

You estimate the product by rounding to the nearest 10:

9 becomes 10 and 54 becomes 50

9 x 50 = 450

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3 years ago

If r = 8 units and h = 5 units, what is the volume of the cylinder shown above? Use 3.14 for . A. 1,607.68 cubic units B. 653.12

soldier1979 [14.2K]

Answer:

<h2>C. 1,004.8 cubic units</h2>

Step-by-step explanation:

$V = ?\\r = 8\\h =5 \\\pi = 3.14\\\\V = \pi r^2 h\\\\V = 3.14 \times 8^2 \times 5\\V = 1004.8 cubic units$

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What is the straight-line distance between (2, -4) and (-3, 8) on a coordinate grid?

Scrat [10]

It is most likely 14 units so it would be A. I think.

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