$(x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}$

4 0

3 years ago

3x-8= 4

BabaBlast [244]

I think the answer is D but i could be wrong

4 0

2 years ago

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consider these graphs: Eric thinks the equation y=(x-3)^2 will match the blue graph. Jenney thinks it will match the red graph.

Dimas [21]

Answer: Jenny

Step-by-step explanation:

By subtitue Y = 0

0=(x-3)^2

And you get 3

3 0

3 years ago

Show that det(A) = 0 without directly evaluating the determinant. A = [-4 1 1 1 1; 1 -4 1 1 1; 1 1 -4 1 1; 1 1 1 -4 1; 1 1 1 1 -

Anuta_ua [19.1K]

You could interpret them as system of equation and answer the question easily

from A

-4x + y + x + w =1

write in same form from A

as this system has five variables and four of them are unknowns it means the unique solution is not possible

So, det (A) = 0

8 0

2 years ago

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