Answer:
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be:
H would decrease either way because h is a variable for a certain number. that number would usually be divided from both sides. you would be eliminating x from the equation. 125x/h. by dividing you are creating a smaller number. by dividing you are decreasing. you're welcome. :)
Answer: I think the answer would be 4+2n=60.. might be wrong though
Step-by-step explanation:
Answer:
h(t) = 16t changes 16 units every t. in the interval t=7 to t=2,
if its asking for:
then it's
16(2) - 16(7) = 32 - 112 = -80