Find an equation of the plane. the plane that passes through the point (−1, 1, 2) and contains the line of intersection of the planes x + y − z = 2 and 4x − y + 5z = 3

Step-by-step explanation:

The plane containing line of intersection of two planes x+y-z-2=0 and 4x-y+5z-3 =0

will be of the form

x+y-z-2+k(4x-y+5z-3) =0

Now given that the plane passes through (-1,1,2)

Substitute to get

-1+1-2-2 +k(-4-1+10-3) =0

-4+2k =0

Or k =2

Substitute in the equation of the plane for k to get real equation.

x+y-z-2+2(4x-y+5z-3) =0

Simplify to get

9x-y+9z-8 =0 is the equation of the required plane.

i pretty sure it is

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: