Find the cosine of the angle between the planes 4x−1y+2z=−4 and the plane 1x+1y−4z=3.

Try this solution:

1. Given: α:4x-y+2z=-4 and β:x+y-4z=3.

if vector n₁⊥α and n₂⊥β, then required cosine is:

where n₁*n₂=4*1+(-1)*1+2*(-4) and |n₁|*|n₂|=√(4²+1²+2²)*√(1²+1²+4²).

2. Using the formula of cosine:

≈0.2572

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option