Answer:
10.50°C
Step-by-step explanation:
Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3
The rate of change in temperature of the bug path can be expressed using the composite formula:
dT/dt = Tx(dx/dt) + Ty(dy/dt)
If x = 2+t; dx/dt = 1
If y = 1+12t; dy/dt = 1/2
Substituting the parameters gotten into dT/dt we will have;
dT/dt = 9(1)+3(1/2)
dT/dt = 9+1.5
dT/dt = 10.50°C/s
Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s
Answer:
Step-by-step explanation:
Perimeter is
P = 2L + 2W. We are given the perimeter as 180 feet, so
180 = 2L + 2W. Solve this for either L or W. I chose W, no reason...
180 - 2L = 2W so
90 - L = W Hold that thought. We'll come back to it in a minute.
Area is
A = LW. We are given the area as 1800 square feet, so
1800 = LW. Sub in 90 - L for W:
1800 = L(90 - L). Distribute to get a quadratic:

Get everything on one side and solve for the length by factoring:

Factor this however you like to factor quadratics, to get the length of
L = 30 and L = 60. First off, the length is longer than the width in general, so if we want to solve for the width, plug in 60 as L in the equation in bold print:
90 - 60 = W and
30 = W
So L = 60 and W = 30
Answer:
C) ![x^{27} (\sqrt[3]{y} )](https://tex.z-dn.net/?f=x%5E%7B27%7D%20%28%5Csqrt%5B3%5D%7By%7D%20%29)
Step-by-step explanation: