1a) A = 4πpw
/4πw = /4πw
A / 4πw = p
1b) A = 4πpw
22 = 4πp(2)
p = 11/4π (≈0.87)
2a) P = 2πr + 2x
P - 2x = 2πr
/2π /2π
P-2x / 2π = r
2b) P = 2πr + 2x
440 = 2πr + 2(110)
r = 110/π (≈35.014)
The smallest region that can be photographed is
5*7 = 35 square km
so we want to know how long it takes to zoom out to an area of
35 * 5 = 175 square km.
Notice that the length and width increase at the same rate, 2 km/sec, and they start out with a difference of two, so when the width has increased from 5 to x, the length will have increased from 7 to x+2. At the desired coverage area, then,
x(x+2) = 175
x^2 + 2x = 175
x^2 + 2x + 1 = 176 [completing the square]
(x+1)^2 = 176
x+1 = ±4√11
x = -1 ±4√11
Since a negative value for x is meaningless here, x must be
-1 + 4√11 = about 12.27 km
and the time it took to increast to that value was
(12.27 - 5) km / (2 km/sec) = about 3.63 seconds
Answer:
28
Step-by-step explanation:
if (AB and DC) parallels lines than the equation is alternate interior and therefore congruent to each other