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

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6