Triangle PQR hAS vertices P (4,-1),Q(-2,7), and R (9,9). find an equation of the median from R

1 answer:

7 0

A median intersects at the midpoint of the opposite length. The midpoint is:

x,m = (4+-2)/2 = 1
y,m = (-1+7)/2 = 3

The midpoint is at (1,3). With this point and point R(9,9), the equation would be:

y = mx + b, where
m = (9 - 3)/(9 - 1) = 0.75
b is the y-intercept
Substituting any point,
3 = 0.75(1)+b
b = 2.25

Thus, the equation for the median is:

y = 0.75x + 2.25

You might be interested in

Please help me with this one

sp2606 [1]

I think 50 oz less
hope it helps

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