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mylen [45]
3 years ago
7

1 over 2 added to 2 over 5

Mathematics
2 answers:
sattari [20]3 years ago
6 0
1 over 2 added to 2 over 5
 For this case, the first thing we must do is rewrite the expression:
 (1/2) + (2/5)
 Now add the fractions taking into account the cross product:
 (1/2) + (2/5) =
 ((5) + (4)) / (10) =
 Finally, we add the values that are in the numerator to obtain the final result:
 (9/10)
 Answer: 
 (1/2) + (2/5) = (9/10)
Natalka [10]3 years ago
4 0
1/2 + 2/5 = 5/10 + 4/10 = 9/10
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Take '1' away from one of them and give it to the other one.

You haven't changed their sum, but now they're<em>  -12</em>  and  <em>-10 </em>.


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3 years ago
Find the indefinite integral. (Use C for the constant of integration.)
mart [117]

Answer:

\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C

tep-by-step explanation:

In order to find the integral:

\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx

we can do the following substitution:

Let's call

u=(x^3-x^2+x)

Then

du = (3x^2-2x+1) dx

which allows us to do convert the original integral into a much simpler one of easy solution:

\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx  = \int\ {u^7 \, du = \frac{1}{8} \,u^8 +C

Therefore, our integral written in terms of "x" would be:

\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C

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3 years ago
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