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4 years ago

Explain how to use proportional reasoning to find 35% of 600

1 answer:

8 0

$\frac{35}{100} = \frac{x}{600}$

This is the equation we would use for this problem.

Now when we cross multiply, we get:
35(600)=100x

Simplify:
21000=100x

Divide each side by 100 to get x b itself:
210=x

So, as you can see, 210 is 35% of 600

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3 years ago

The sum of three consecutive integers is 189. What are the integers?

Leto [7]

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Step-by-step explanation:

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8 0

4 years ago

Complete each statement then state the property or properties illustrated by each statement on

VMariaS [17]

Answer:

(8n)(-77)=-77(n*8)

We move all terms to the left:

(8n)(-77)-(-77(n*8))=0

We add all the numbers together, and all the variables

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Back to the equation:

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We get rid of parentheses

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n=0/1

n=0

the property is Associative Property

Step-by-step explanation:

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7 0

3 years ago

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

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

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

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

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

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

3 0

2 years ago