Q: How can you find 2x3?

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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

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

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If x = -7, then x3 = -343 so that 2x3 = -686 y = 2 then 15 y = 30 So 2x3 + 15y = -686 + 30 = -656

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1x2 1x3 1x4 and so on... 2x3 2x4

(2x3)+(3x5)-(3x2)= 2x3=6 3x5=15 3x2=6 So..... 6x25-6= 6x25=150 150+6=156

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

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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no

False

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

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

2x (1+3)=(2x1+(2x3)

1x6 2x3

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