Answer:
Both fireworks will explode after 1 seconds after firework b launches.
Step-by-step explanation:
Given:
Speed of fire work A= 300 ft/s
Speed of Firework B=240 ft/s
Time before which fire work b is launched =0.25s
To Find:
How many seconds after firework b launches will both fireworks explode=?
Solution:
Let t be the time(seconds) after which both the fireworks explode.
By the time the firework a has been launched, Firework B has been launch 0.25 s, So we can treat them as two separate equation
Firework A= 330(t)
Firework B=240(t)+240(0.25)
Since we need to know the same time after which they explode, we can equate both the equations
330(t) = 240(t)+240(0.25)
300(t)= 240(t)+60
300(t)-240(t)= 60
60(t)=60

t=1
Answer:
2.5 years
Step-by-step explanation:
Interest= P×R×T÷100
T= 100I÷P×R
T= 100×400÷8000×2
T= 40000÷16000
T= 5÷2
T= 2.5 years
Answer:
It would be worth $5,040 after 7 years
Step-by-step explanation:
Multiply 40% by 7 and get 280, then turn it into a percent, .280, finally multiply it by $18,000 to get $5,040
Answer:
-69x-66
Step-by-step explanation:
1) given: a₁=-4x-1; a₂=-9x-6; a₃=-14x-11. Find: a₁₄-?
2) a₁₄=a₁+13d, where d - the difference between neighbor-terms;
3) d=a₂-a₁; ⇔ d= -9x-6+4x+1= -5x-5;
4) a₁₄= -4x-1+13*(-5x-5)= -69x-66.