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1 year ago

Line c passes through points (1, 14) and (7, 7). Line d is perpendicular to c. What is the slope of line d?

Mathematics

1 answer:

anyanavicka [17]1 year ago

3 0

Answer:

6/7

Step-by-step explanation:

Your first step will be to find the slope of Line C. Using the slope formula of $\frac{y_{2}-y_1 }{x_2-x_1}$ , the slope of Line C is $\frac{7-14}{7-1} = \frac{-7}{6}$ .

If Line D is perpendicular to C, that means the slope is the negative reciprocal of Line C. So, that means the slope is 6/7.

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

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