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3 years ago

What is (2+y)³ written out? 8 + 12y + 6y² + y³ 6 + 8y + 4y² + y³

2 answers:

7 0

$\boxed{(2+y)^3-cube\ the\ sum\ of\ the\ numbers\ 2\ and\ y}$

$Use:\ (a+b)^3=(a+b)(a+b)(a+b)=a^3+3a^2b+3ab^2+b^3\\----------------------------\\(2+y)^3=2^3+3\cdot2^2\cdot y+3\cdot2\cdot y^2+y^3=8+3\cdot4y+6y^2+y^3\\\\=8+12y+6y^2+y^3\\========================================$

$8+12y+6y^2+6y^3+8y+4y^2+y^3\\=8+(12y+8y)+(6y^2+4y^2)+(6y^3+y^3)\\=8+20y+10y^2+7y^3$

geniusboy [140]3 years ago

6 0

(2+y)3

=(2+y)*(2+y)*(2+y)

=(2+y)(y2+4y+4)

=(2)(y^2)+(2)(4y)+(2)(4)+(y)(y^2)+(y)(4y)+(y)(4)

=2y^2+8y+8+y^3+4y^2+4y

=y^3+6y^2+12y+8

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8 + 12y + 6y² + y³ 6 + 8y + 4y² + y³

Combine Like Terms:

=8+12y+6y^2+6y^3+8y+4y^2+y3

=(6y^3+y^3)+(6y^2+4y^2)+(12y+8y)+(8)

=7y^3+10y^2+20y+8

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Complete Question:

Find both the vector equation and the parametric equations of the line through (0,0,0) that is perpendicular to both $u = < 2, 0, 2>$ and $w = < -2, 1, 0>$ where t = 0 corresponds to the first given point.

Answer:

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$u \times w = \left[\begin{array}{ccc}i&j&k\\2&0&2\\-2&1&0\end{array}\right] \\\\u \times w = i(0-2) -j(0+4) + k(2)\\\\u \times w = -2i - 4j + 2k\\\\u \times w = < -2, -4, 2>$

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