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levacccp [35]
3 years ago
11

Simplify 4x (2x^2-3x-4)

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

8x^3 -12x^2 - 12x

Step-by-step explanation:

Break all the brackets and you would got:

4x*(2x^2-3x-4)

= 4x*2x^2 - 4x*3x - 4x*4

= 8x^2+1 - 12x^2 - 12x

= 8x^3 -12x^2 - 12x

This would be the simplified form of the equation

Hope this help you :3

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Step-by-step explanation:

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Given f(x) = 3x - 1 and g(x)= -x + 6, find f(-2) + g(5).<br> A. -6<br> B. 6<br> C. 8
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Answer:

A. -6

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Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straig
Brilliant_brown [7]

Stokes' theorem equates the line integral of \vec F along the curve to the surface integral of the curl of \vec F over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it T) by

\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)

\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to T to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of T is traversed in the counterclockwise direction when viewed from above.

Compute the curl of \vec F:

\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)

Then by Stokes' theorem,

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\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

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3 years ago
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