The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
(c) III
Step-by-step explanation:
If you simplify the equations and the left side is identical to the right side, then there are an infinite number of solutions: the equation is true for all values of x.
Another way to simplify the equation is to subtract the right side from both sides. If that simplifies to 0 = 0, then there are an infinite number of solutions.
__
<h3>I. </h3>
2x -6 -6x = 2 -4x . . . . eliminate parentheses
-4x -6 = -4x +2 . . . . no solutions (no value of x makes this true)
__
<h3>II.</h3>
x +2 = 15x +10 +2x . . . . eliminate parentheses
x +2 = 17x +10 . . . . one solution (x=-1/2)
__
<h3>III.</h3>
4 +6x = 6x +4 . . . . eliminate parentheses
6x +4 = 6x +4 . . . . infinite solutions
__
<h3>IV.</h3>
6x +24 = 2x -4 . . . . eliminate parentheses; one solution (x=-7)
Answer:
5tyg5frhhu
Step-by-step explanation:
fgrhdwjeygfrrredfr