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3 years ago

Plssss I need help Im in classsss plsss

Mathematics

1 answer:

Sav [38]3 years ago

5 0

Step-by-step explanation:

Y=-4x+3

2x+3y=19

2x+3(-4x+3)=19

2x-12x+9=19

-10x=10

X=-1

y=-4*(-1)+3=4+3=7

x=-1 y=7

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Which is a better price: 5 for $1.00, 4 for $0.85, 2 for $0.25, or 6 for $1.10?

Angelina_Jolie [31]

Answer:

2 for $0.25

Step-by-step explanation:

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4 years ago

Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)

Amanda [17]

Answer:

$f_x(x,y,z)=4\sin (y-z)$

$f_x(x,y,z)=4x\cos (y-z)$

$f_z(x,y,z)=-4x\cos (y-z)$

Step-by-step explanation:

The given function is

$f(x,y,z)=4x\sin (y-z)$

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

$f_x(x,y,z)=4(1)\sin (y-z)$

$f_x(x,y,z)=4\sin (y-z)$

Differentiate partially w.r.t. y and x, z are constants.

$f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)$

$f_y(x,y,z)=4x\cos (y-z)$

Differentiate partially w.r.t. z and x, y are constants.

$f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)$

$f_z(x,y,z)=4x\cos (y-z)(-1)$

$f_z(x,y,z)=-4x\cos (y-z)$

Therefore, the first partial derivatives of the function are $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .

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skelet666 [1.2K]

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+7 +7
---------------------
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----------------------
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*3 *3
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Hope this helps! :)

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