A business owner opens one store in town A. The equation p(x) = 10,000(1.075)' represents the anticipated profit aft- t years. The business owner opens a store in town B six months later and predicts the profit from that store to increas
at the same rate. Assume that the initial profit from the store in town B is the same as the initial profit from the store ir
town A. At any time after both stores have opened, how does the profit from the store in town B compare with the pro
from the store in town A?
a
O 65%
O 96%
104%
154%
Mark this and retium

96%

Step-by-step explanation:

he rigth equation to anticipate the profit after t years is p(t) = 10,000 (1.075)^t

So, given that both

store A and store B follow the same equations but t is different for them, you can right: Store A: pA (t)  10,000 (1.075)^t

Store B: pB(t'): 10,000 (1.075)^t'

=> pA(t) / pB(t') = 1.075^t / 1.075^t'

=> pA(t) / pB(t') = 1.075 ^ (t - t')

And t - t' = 0.5 years

=> pA(t) / pB(t') = 1.075 ^ (0.5) = 1.0368

or pB(t') / pA(t) = 1.075^(-0.5) = 0.964

=> pB(t') ≈ 0.96 * pA(t)

This means that the profit of store B is about 96%of  the profit of store A at any time after both stores have opened.

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

step-by-step explanation: i dont