Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r = 6.
(Enter the dimensions as a comma-separated list.)

Explanation:

Given:-

Initial velocity (u) = 10m/s

Acceleration (a) = 0.2m/s²

Time (t) = ½ minute

To Find:-

The distance traveled by the car.

The final velocity of the car.

Solution:-

u = 10m/s , a = 0.2m/s²

Time (t) = ½ minute = 30 sec

Final velocity (v) = ?

Distance (s) = ?

Using 1st equation of motion:-

\sf \implies v = u + at⟹v=u+at

\sf \implies v = 10 + (0.2 \times 30)⟹v=10+(0.2×30)

\sf \implies v = 10 + 6⟹v=10+6

\sf \implies v = 16m/s⟹v=16m/s

\sf \therefore \underline{\:\: \underline{\: Final \: velocity = 16m/s\:}\:\:}∴

Finalvelocity=16m/s

Using 3rd equation of motion:-

\sf \implies s = ut + \dfrac{1}{2}at^2⟹s=ut+

2

1

at

2

\sf \implies s = (10 \times 30) + \dfrac{1}{2} \times 0.2 \times {(30)}^{2}⟹s=(10×30)+

2

1

×0.2×(30)

2

\sf \implies s = (300) + (0.1 \times 900)⟹s=(300)+(0.1×900)

\sf \implies s = 300+ 90⟹s=300+90

\sf \implies s = 390⟹s=390

\sf \therefore \underline{\:\: \underline{\: Distance \: travelled = 390m\:}\:\:}∴

Distancetravelled=390m

Distance  travelled = 390m

Distance travelled=390m

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

ur mom

step-by-step explanation: