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3 years ago

What is the value of the coefficient of the fourth term of the expansion of (a+b)^5 ?

Mathematics

1 answer:

melomori [17]3 years ago

6 0

Answer:

I don't know if you wanted it expanded using "Pascal's Triangle" or "Binomial Theorem" but either way is the same. Your expanded equation is:

a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

I hope this is any help to you, have a nice day.

Step-by-step explanation:

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Cerrena [4.2K]

Answer:

Step-by-step explanation:

As x increases four units from -4 to 0, y decreases four units from 0 to -4

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Consider the surface f (x comma y comma z )f(x,y,z)equals=negative 2 x squared plus 2 y squared minus 3 z squared plus 3 equals

trapecia [35]

Answer:

a) (8,8,-6)

b) 4x+4y+3z = -3

Step-by-step explanation:

The surface is given by the equation

f(x,y,z) = 0 where

$f(x,y,z)=-2x^2+2y^2-3z^2+3$

The gradient of this function is the vector

$(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})=(-4x,4y,-6z)$

If we evaluate it in the point P = (-2,2,1) we obtain the point

(8,8,-6)

The vectors with their tails at P are of the form

(-2,2,1)-(x,y,z) = (-2-x, 2-y, 1-z)

as they must be orthogonal to the gradient, they must be orthogonal to the vector (8,8,6) so their inner product is 0

$(-2-x,2-y,1-z)\cdot(8,8,6)=0\Rightarrow -16-8x+16-8y+6-6z=0\Rightarrow 4x+4y+3z=-3$

and the equation of the desired plane is

4x+4y+3z = -3

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3 years ago