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Q: How do rewrite 2x3 plus 5?

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f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this

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your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i

(x + 1) / (2x3 + 3x + 5) --- Can't depict long division here, but you can work this out by seeing if (x + 1) is a factor of (2x3 + 3x + 5). If it is, then the answer will be 1/(the other factor). In this case, that is true. (2x3 + 3x + 5)/(x + 1) = 2x2 - 2x + 5, so the term term above is equal to: 1 / (2x2 - 2x + 5)

2x(x + 5)(x - 2)

Since there is nothing following, none of them!

(x + 5)2

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2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7

10x3 + 3x2

103

It depends on whether or not there are parentheses around the 2x3. If yes, the answer is 26. If not, the answer is 48.

6

Yes, you can rewrite a cuda program originally written in c in c plus plus.

4X2+X3+3X+7Y+2X3 can only be simplified to 3X3+4X2+3X+7Y, because X3 and 2X3 are the only like terms.

Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)

Six.

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Not necessarily.

4(x + y - 5) = 0

What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).

Quotient: 2x3-x2-14x+42 Remainder: -131 over (x+3)

== == y = -x + 5

-2+5 Addition is commutative, so you can rewrite this as 5-2 5-2=3

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