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LiRa [457]
3 years ago
13

Find the work done by force dield f(x,y)=x^2i+ye^xj on a particle that moves along parabola x=y^2+1 from(1,0) to (2,1)

Mathematics
1 answer:
NemiM [27]3 years ago
8 0
Call the parabola P, parameterized by \mathbf r(y)=\langle y^2+1,y\rangle with 0\le y\le 1\rangle. Then the work done by \mathbf f(x,y) along P is

\displaystyle\int_P\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{y=0}^{y=1}\mathbf f(x(y),y)\cdot\dfrac{\mathrm d\mathbf r(y)}{\mathrm dy}\,\mathrm dy
=\displaystyle\int_0^1\langle(y^2+1)^2,ye^{y^2+1}\rangle\cdot\langle2y,1\rangle\,\mathrm dy
=\displaystyle\int_0^1(2y^5+4y^3+2y+ye^{y^2+1})\,\mathrm dy=\frac73-\frac e2+\frac{e^2}2
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Let f(x) = 3 * sqrt(x) + 7. Define g$to be the inverse of f.
aliina [53]

By using the definition of inverse functions, we will see that:

g(g(f(f(f(36))))) = f(36) = 25.

<h3>What are inverse functions?</h3>

Two functions f(x) and g(x) are inverses if:

f( g(x) ) = x

g( f(x) ) = x

Then we can rewrite:

g(g(f(f(f(36)))))

First, we can see that:

g(f(f(f(36)))) = f(f(36))

Replacing that in our expression, we get:

g(g(f(f(f36))))) = g(f(f(36)))

And the above expression is equal to f(36), to be sure of that, let's replace:

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Then we can rewrite:

g(f(f(36))) = g(f(u))

And by definition, the above thing is equal to u:

g(f(u)) = u = f(36).

Finally, we conclude that:

g(g(f(f(f(36))))) = f(36) = 3*√36 + 7 = 3*6 + 7 = 25

If you want to learn more about inverse functions, you can read:

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4 0
2 years ago
PLEASE HELP 25 POINTS!!
eduard

Answer:

(27 y^(-8))/(4 x^6)

Step-by-step explanation:

Simplify the following:

(4 (3 x^2 y^4)^3)/(2 x^3 y^5)^4

Multiply each exponent in 2 x^3 y^5 by 4:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^(4×5))

4×5 = 20:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^20)

4×3 = 12:

(4 (3 x^2 y^4)^3)/(2^4 x^12 y^20)

2^4 = (2^2)^2:

(4 (3 x^2 y^4)^3)/((2^2)^2 x^12 y^20)

2^2 = 4:

(4 (3 x^2 y^4)^3)/(4^2 x^12 y^20)

4^2 = 16:

(4 (3 x^2 y^4)^3)/(16 x^12 y^20)

Multiply each exponent in 3 x^2 y^4 by 3:

(4×3^3 x^(3×2) y^(3×4))/(16 x^12 y^20)

3×4 = 12:

(4×3^3 x^(3×2) y^12)/(16 x^12 y^20)

3×2 = 6:

(4×3^3 x^6 y^12)/(16 x^12 y^20)

3^3 = 3×3^2:

(4×3×3^2 x^6 y^12)/(16 x^12 y^20)

3^2 = 9:

(4×3×9 x^6 y^12)/(16 x^12 y^20)

3×9 = 27:

(4×27 x^6 y^12)/(16 x^12 y^20)

4/16 = 4/(4×4) = 1/4:

(27 x^6 y^12)/(4 x^12 y^20)

Combine powers. (27 x^6 y^12)/(4 x^12 y^20) = (27 x^(6 - 12) y^(12 - 20))/4:

(27 x^(6 - 12) y^(12 - 20))/4

6 - 12 = -6:

(27 x^(-6) y^(12 - 20))/4

12 - 20 = -8:

Answer: (27 y^(-8))/(4 x^6)

5 0
3 years ago
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