A pyrotechnician plans for two fireworks to explode together at the same height in the air. They travel at speeds shown below. Firework B is launched 0.25 s before Firework A. How many seconds after Firework B launches will both fireworks explode?

Firework A

360 ft/s

Firework B

340 ft/s

both fireworks will explode 4.5 seconds after  Firework B launches

Step-by-step explanation:

Given;

speed of firework A, = 360 ft/s

speed of firework B, = 340 ft/s

If the two fireworks explodes at the same height, then the height attained by the two fireworks are equal.

let the distance traveled by each firework before explosion = d

Distance = speed x time

Distance A = Distance B

speed A x time = speed B x time

let the time both fireworks explodes after Firework B launches = t

() t = ( ) t

360t = 340t

if firework B is launched 0.25 s before Firework A, for the time of the two fireworks to be equal since we are considering time (t) after 0.25 seconds, we will have;

360(t-0.25) = 340t

360t  - 90 = 340t

360 t - 340 t = 90

20 t = 90

t = 90/20

t = 4.5 seconds

Therefore, both fireworks will explode 4.5 seconds after  Firework B launches

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.