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Vinvika [58]
3 years ago
14

(2a^2b + 3ab^2 - b^3) - (4b^3 - 3ab^2 -6a^2b)

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
7 0

<em>Hi there!</em>

<em>~</em>

<em></em>(2a^2b+3ab^2-b^3)-(4b^3-3ab^2-6a^2b)<em></em>

<em>Distribute the Negative Sign:</em>

<em></em>= 2a^2 b + 3ab^2 - b^3 + -1(4b^3 - 3ab^2 - 6a^2b)\\= 2a^2 b + 3ab^2 + - b^3 + -1 (4b^3) + -1 (-3ab^2) + -1 (-6a^2b)\\= 2a^2 b + 3ab^2 + -b^3 + -4b^3 + 3ab^2 + 6a^2 b<em></em>

<em>Combine Like Terms:</em>

<em></em>= 2a^2 b + 3ab^2 - b^3 + -4ab^3 + 3ab^2 + 6a^2 b\\= (2a^2 b + 6a^2 b) + ( 3ab^2 + 3ab^2) + (-b^3 + -4b^3) \\= 8a^2 b + 6ab^2 + -5b^3<em></em>

<em>Answer : </em>8a^2 b + 6ab^2 + -5b^3<em></em>

<em>Hope this helped you!</em>

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Answer:

\bold{x = 34^\circ}

Step-by-step explanation:

Given:

cos x = sin (x + 22^\circ)

To solve:

The given equation.

Solution:

First of all, let us consider an important property of sine and cosine.

sin(90^\circ-\theta )=cos\theta

OR

cos(90^\circ-\theta )=sin\theta

We can apply above property to solve for x as per given equation.

cos x = sin (x + 22^\circ)

Changing cosx to sine form:

cosx=sin(90^\circ-x)

cosx=sin(90^\circ-x) = sin(x+22^\circ)

\therefore 90^\circ-x=x+22^\circ\\\Rightarrow 90^\circ-22^\circ=x+x\\\Rightarrow 2x=68^\circ\\\Rightarrow \bold{x = 34^\circ}

So, solution to the equation cos x = sin (x + 22^\circ) is:

\bold{x = 34^\circ}

7 0
3 years ago
So can someone explain how the heck does 4 x 4 x 4 = 64??
sweet-ann [11.9K]

Answer:

it is more like, 4x4=16

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Step-by-step explanation:

I know it can be confusing, but you will eventually get it :)

i think you were adding 4+4+4=12, but that is not what you are supposed to do here.

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2 years ago
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Step-by-step explanation:

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9x16 power 3 =36864

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