Im guessing either 2/3*570 or 1/3*5700
Answer: Hello there!
this type of equations in one dimension (when all the factors are constants) are written as:
h = initial position + initial velocity*t + (acceleration/2)*t^2
First, let's describe the hunter's equation:
We know that Graham moves with a velocity of 1.5 ft/s, and when he is 18 ft above the ground, Hunter throws the ball, and because Graham is pulled with a cable, he is not affected by gravity.
If we define t= 0 when Graham is 18 ft above the ground, the equation for Graham height (in feet) is:
h = 18 + 1.5t
where t in seconds.
Now, the equation for the ball:
We know that at t= 0, the ball is thrown from an initial distance of 5ft, with an initial velocity of 24ft/s and is affected by gravity acceleration g, where g is equal to: 32.2 ft/s (notice that the gravity pulls the ball downwards, so it will have a negative sign)
the equation for the ball is:
h = 5 + 24t - (32.2/2)t^2 = 5 + 24t - 16.1t^2
So the system is:
h = 18 + 1.5t
h = 5 +24t - 16.1t^2
so the right answer is A
Answer:
6in
Step-by-step explanation:
Given data
Volume of the lockbox= 312 cubic inches
Height= 4in
Length= 13in
Width= ???
The expression for the volume of box is
V= L*W*H
substitute
312=13*W*4
312=52W
W= 312/52
W=6 in
Hence the width is 6in
Answer:
it's d it makes alot of sense and ik it's right
