Water drips from the nozzle of a shower onto the floor 200 cm below. the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. when the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

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!

the 4th choice is the correct one

step-by-step explanation:

i dont know sorry

step-by-step explanation: