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2 years ago

Please help i will give brainliest

Mathematics

2 answers:

zhannawk [14.2K]2 years ago

8 0

Answer:

-4.2, -3.5, -2.1, -1.5, -1, -0.5, 0.5, 3.5, 4.8

Step-by-step explanation:

jarptica [38.1K]2 years ago

5 0

Answer:

-4.2

-3.5,

-2.1,

-1

-1.5

-0.5

3.5,

4.8

Step-by-step explanation: Please feel free to point out if I'm wrong or if I missed any (i did spaces so you could see easier, also from top to bottom is least to greatest)

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1 year ago

Read 2 more answers

Plz help I need the answer with steps

defon

Now subtract 3 from 6

8 0

2 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 $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .