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Montano1993 [528]
3 years ago
11

I need some Help please

Mathematics
1 answer:
djverab [1.8K]3 years ago
6 0

Answer:

a) f(2) = 4

b) g(3) = 3

c) g(1/3) = -1

d) f(3) = 1/2

e) g(-2) = 2

f) f(π) = 3

Step-by-step explanation:

We are given:

f={(2,4),(-4,3),(3,1/2),(π,3)} and g={(3,3),(-2,2),(1/3,-1)}

We need to find the following values of f and g

a) f(2)

f(2) means when x=2, we need to find the value of y.

Looking at the given f={(2,4),(-4,3),(3,1/2),(π,3)} , when x =2 then y=4

So, f(2) = 4

b) g(3)

g(3) means when x=3, we need to find the value of y.

Looking at the given g={(3,3),(-2,2),(1/3,-1)}, when x =3 then y=3

So, g(3) = 3

c) g(1/3)

g(1/3) means when x=1/3, we need to find the value of y.

Looking at the given g={(3,3),(-2,2),(1/3,-1)}, when x =1/3 then y=-1

So, g(3) = -1

d) f(3)

f(3) means when x=3, we need to find the value of y.

Looking at the given f={(2,4),(-4,3),(3,1/2),(π,3)} , when x =3 then y=1/2

So, f(3) = 1/2

e) g(-2)

g(-2) means when x=-2, we need to find the value of y.

Looking at the given g={(3,3),(-2,2),(1/3,-1)}, when x =-2 then y=2

So, g(-2) = 2

f) f(π)

f(π) means when x=π, we need to find the value of y.

Looking at the given f={(2,4),(-4,3),(3,1/2),(π,3)} , when x =π then y=3

So, f(π) = 3

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Step-by-step explanation:

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\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Step-by-step explanation:

Given equation of ellipsoids,

u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}

The vector normal to the given equation of ellipsoid will be given by

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            =\bigtriangledown u

           

=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})

           

=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}

           

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Hence, the unit normal vector can be given by,

\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}

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=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

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