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3 years ago

How to solve this in simplified form.

Mathematics

2 answers:

nika2105 [10]3 years ago

7 0

When multiplying exponents, add them
e^10*e^-2=e^(10-2)=e^8

kaheart [24]3 years ago

7 0

I think you are aware of the power rules

I attached charts that would help you

For your case:
$\Large e^{10}*e^{-2}=e^{(10)+(-2)}=e^{10-2}=e^8

$

I hope that helps!

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Will the square of a negative number always be positive? Explain why.

qwelly [4]

the square of a negative number always be positive

eg: -2² = -2×-2 = (-×- = +) +4

the cube of a negative number always be positive

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hope it helps...!!!

6 0

2 years ago

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The ratio of the length and the breadth of rectangle is 2:1.If the rectangle is 12m long, find the its breadth

Vlad1618 [11]

Answer:

Step-by-step explanation:

The ratio of the length and the breadth of rectangle=2:1

Let length be 2x, then breadth = x

Now,

Length(l)=12 m

or,2x=12m

or,x=6m

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4 0

2 years ago

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Ainat [17]

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3 years ago

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Lelechka [254]

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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:

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

7 0

3 years ago