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:
the number is 4
Step-by-step explanation:
we can write this as
x + 15 = 19
now we transpose
x= 19-15
x= 4
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Answer:
x + 49 + 28 = 180
Step-by-step explanation:
The angles on a straight line with x° are 28° and 49° respectively, side by side with x°. This is because, vertical angles are equal. Therefore, the angle vertically opposite 28° and 49° are equal to 28° and 49° respectively.
Therefore, since angles on a straight line is 180°, thus:
x + 49 + 28 = 180
