Acar leaving a stop sign accelerates constantly from a speed of 0 feet per second to reach a speed of 44 feet per second. the distance of the car from the stop sign, d d , in feet, at time t t , in seconds, can be found using the equation d=1.1 t 2 d=1.1t2 . what is the average speed of the car, in feet per second, between t=2 t=2 and t=5 t=5 ? 5.5 6.6 7.7 8.5

step-by-step explanation:

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation:

answer:

at time x=0 we have 1000, at time x=1 we have 2000, at time x=2 we have 4000.  

so that's an initial value of 1000 and a ratio of 2,

y = 1000 (2^x)

hope this !