Let t be the plane-2x-2y+z =-13. find the shortest distance d from the point po=(-5,-5,-3) to t, and the point q in t that is closest to po. use the square root symbol 'where needed to give an exact value for your answer. d 0 0 (0, 0, 0)

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane , then r will have the next parametric equations:

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

Substitute the value of in the parametric equations:

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.