Answer:
The answer is C.
Step-by-step explanation:
Given formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height.
In this case, the initial postion is a platform 30ft above ground so h0=+30
The initial velocity is 38 ft/s straight up into the air so v0=+38
h(t)=-16t2+38t+30
When the object hits the ground, h=0.
h=-16t2+38t+30=0
Simplifying 8t2-19t-15=0
(8t+5)(t-3)=0
t=-5/8 or 3
As time cannot be -ve, t=3s. The answer is C.
To solve, we will follow the steps below:
3x+y=11 --------------------------(1)
5x-y=21 ------------------------------(2)
since y have the same coefficient, we can eliminate it directly by adding equation (1) and (2)
adding equation (1) and (2) will result;
8x =32
divide both-side of the equation by 8
x = 4
We move on to eliminate x and then solve for y
To eliminate x, we have to make sure the coefficient of the two equations are the same.
We can achieve this by multiplying through equation (1) by 5 and equation (2) by 3
The result will be;
15x + 5y = 55 ----------------------------(3)
15x - 3y =63 --------------------------------(4)
subtract equation (4) from equation(3)
8y = -8
divide both-side of the equation by 8
y = -1
Answer:
See below.
Step-by-step explanation:
Sin (4π/3) = -sqrt(3)/2
Cos(4π/3) = -1/2
Answer:
<em>All real solutions.</em>
Step-by-step explanation:
5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13
-15x-10-x+3=-16x-20+13
-16x-7=-16x-7
All real solutions.
hope this halps... :D lol
