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3 years ago

PLEASE HELP THIS IS A TEST

Mathematics

1 answer:

cluponka [151]3 years ago

8 0

Answer:

C, F, and maybe D are correct.

Step-by-step explanation:

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Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

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

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

4 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Math Help? Please! : )

Ray Of Light [21]

1/2 is the probability, or 50%. There are 4 teams of S and B n the Y, 2 are B. so its 2/4

5 0

3 years ago

Factor x^4 + x y^3 + x^3 + y^3 completely. Show your work.

andrey2020 [161]

$\bf x^4+xy^3+x^3+y^3\impliedby \textit{now let's do some grouping}
\\\\\\
(x^4+xy^3)~~+~~(x^3+y^3)\implies x(x^3+y^3)~~+~~(x^3+y^3)
\\\\\\
\stackrel{\stackrel{common}{factor}}{(x^3+y^3)}(x+1)\\\\
-------------------------------\\\\
\textit{now recall that }\qquad \qquad \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\
-------------------------------\\\\
(x+y)(x^2-xy+y^2)(x+1)$

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3 years ago

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Shalnov [3]

Answer: 1. A Please give me a thanks and a brainly

Step-by-step explanation:

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3 years ago