A square plate RequalsStartSet (x comma y ): 0 less than or equals x less than or equals 1 comma 0 less than or equals y less than or equals 1 EndSet has a temperature distribution Upper T (x comma y )equals100 minus 25 x minus 40 y.

a. Sketch two level curves of the temperature in the plate.
b. Find the gradient of the temperature gradient Upper T (x comma y ).

Step-by-step explanation:

We are given the following:

and

a). Recall that a level curve of a function f(x,y) is given by where c is a constant. That is, all the points in the set of interest to which the function applied to the points is exactly the value c.

Consider c = 80. So we get

which implies that (Graph 1).

We can also consider c=60, which gives us

which implies that . (Graph 2)

b)Recall that the gradient of a function f(x,y) is given by

In this case,

Thus, the gradient of T is given by

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p