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3 years ago

Hepl with polynomials (2/3Y2 - 3/4) + (5/6Y2 + 4)

Mathematics

1 answer:

hodyreva [135]3 years ago

5 0

4y/3 - 3/4 + 5/6y (multiply by) x 2 + 4

4y/3 - 3/4 + 5y/3 + 4

collect like terms: (4y/3 + 5y/3) + ( - 3/4 + 4)

simplify: 3y + 13/4 <----- FINAL ANSWER

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Answer:

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Step-by-step explanation:

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3 years ago

The work W required to move a particle from a far distance to within radius r of another particle varies jointly as the product

tangare [24]

Answer:

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Step-by-step explanation:

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3 years ago

What is the product?<br><br> (6r-1)(-8r-3)

Ilia_Sergeevich [38]

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3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

3 years ago