Let x be the distance (in feet) along the road that the car has traveled and h be the distance (in feet) between the car
and the observer.
(a) Before the car passes the observer, we have dh/dt < 0; after it passes, we have dh/dt > 0. So at the instant it passes the observer we have
dh/dt = 0, given that dh/dt varies continuously since the car travels at a constant velocity.
W= Width
L= Length
2L+2w=52
2W-4=L
Plug in L into the first equation
2(2W-4)+2w=52
4w-8+2w=52
6W=60
W=10 feet
Plug that into the second equation
2(10)-4=L
L=16
Answer:
D
Step-by-step explanation:
sorry idk how to explain it
Answer: See attached table
Step-by-step explanation:
+---+---+---+
| 1 | 8 | 2 |
+---+---+---+
| 6 | 4 | 2 |
+---+---+---+
| 5 | 0 | 7 |
+---+---+---+
<u>Proofs:</u>
First row: 1+6+5 = 12
Second row: 8+4+0 = 12
Third row: 2+2+7 = 12
First column: 1+8+2 = 12
Second column: 6+4+2 = 12
Third column: 5+0+7 = 12
Diagonal starting from top left to bottom right: 1+4+7 = 12
Diagonal staring from top right to bottom left: 2+4+5 = 12
Given that AB = 16, BC = 5, and CD = 3, which is the length of secant DE?It might be useful to see ...