(1 point) Find an equation of the tangent plane to the parametric surface x=5rcosθx=5rcos⁡θ, y=4rsinθy=4rsin⁡θ, z=rz=r at the point (52–√,42–√,2)(52,42,2) when r=2r=2, θ=π/4θ=π/4. z = Note: Your answer should be an expression of x and y; e. g. "3x - 4y"

Denote the given surface by the vector-valued function,

Then the normal vector to this plane is

When and , the normal vector is

and so the equation of the tangent plane is

Judging by the "note", you need to write the equation as , which is trivial:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."