Answer:
2.54 seconds
Step-by-step explanation:
We can use the following equation to model the vertical position of the ball:
S = So + Vo*t + a*t^2/2
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.
Then, using S = 2.5, So = 0.4, Vo = 14 and a = -9.8 m/s2, we have that:
2.5 = 0.4 + 14*t - 4.9t^2
4.9t^2 - 14t + 2.1 = 0
Solving this quadratic equation, we have that t1 = 2.6983 s and t2 = 0.1588 s.
Between these times, the ball will be higher than 2.5 m, so the amount of time the ball will be higher than 2.5 m is:
t1 - t2 = 2.6983 - 0.1588 = 2.54 seconds
Answer:
81°+x=180°[co interior angles]
x=180°-81°
x=99°
x°=(3y+33)°
99°=(3y+33)°
99-33=3y
66=3y
66/3=y
22=y
Answer:
16/13 is the answer.
26-10 is 16 and 2x5+3 is 13
16x + 64
Hope this helps:)
You can write the two equations x = 4y - 7 and x + y = 93; substitute the value of x into the other equation
(4y - 7) + y = 93; combine like terms
5y - 7 = 93; add 7 to both sides
5y = 100; divide both sides by 5
y = 20; substitute this value into the other equation
x + 20 = 93
x = 73
So the numbers are 20 and 73
