Answer:
y=5/2x
Step-by-step explanation:
The line needs to go through the points (0,0) and (2,5).
With y=5/2x, there is no y-intercept, so it starts on the origin, (0,0). With the slope 5/2, you would rise 5 (y) from the origin, and run 2 (x), which gives you the point (2,5).
Answer:
So the given equation is h= -5t^2+40t
the first question asks when it will be 75 meters above the ground. So plug in 75 for h and find t
75= -5t^2+40t
start by dividing everything by 5
15= -x^2 + 8x
then you can add x squared to both sides and subtract 8x
x^2 - 8x + 15 =0
then you can plug it into the quadratic formula, which is a pain to type so I did it on paper.
x = 5 or 3
so it will be 75 meters above the ground at 3 seconds and 5 seconds
Rewriting the equation
ok so I accidentally already factored it when I divided everything by 5...
I believe the answer here is x^2 - 8x + 15 =0
The meaning of the x intercepts is when h=0 or the object is 0 meters above the ground (basically when it starts and when it ends)
the maximum height is greater than 75 because it gets to 75 meters 3 seconds after it is thrown, and continues going higher before it comes back to 75 two seconds later.
The object is airborne for 8 seconds because the second x intercept is at (8,0)
Ah, good ol' substitution. You need to take the phrase y+2 and replace every x with it. Then solve for y.
y=(y+2)^3-4(y+2)^2-6(y+2)+8
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