I don't know for sure, but I think there might be an error in the way you copied your function. In all of my dealings with these types of problems, I have learned the formula to be h(t) = 15 + 10t - 16t^2. Notice the plus in front of the "10t". This is due to the fact that if he pushes off of the overhang he would have an upward force of 10 feet per second as soon as his feet left the ground. The only thing pulling him back to Earth is gravity, modeled by the "-16t^2". This is derived from a bit of slightly advanced physics involving the gravitational constant, but let's work under my formula for a second...
Either way, we will wind up using the Quadratic Formula (or possibly factoring if the numbers are easy enough to work with). So let's start.
h(t)= 15 +10t - 16t^2
In order to use the QF or factoring I will need to make h(t)=0. Simply done by:
0= 15 +10t -16t^2
Looking at the numbers, I'd prefer to use the QF so here it is:
I know that my answer will need to be positive since you can't have a negative value when dealing with time, so I will eliminate the positive sign from the "+or-" part leaving me with:
And I know that a= -16, b= 10, and c=15. So all that's left to do is substitute and solve.
There's a decent amount of math that would be difficult and sloppy for me to do over the computer, but all you need to do is solve the rest of the equation and you would get your answer.
Exact answer:
or rounded answer: ≈1.31 seconds.
Hope this helps!
NoThisIsPatrick
Answer:
the answer for this question would be 28
Step-by-step explanation:
Answer:
26
Step-by-step explanation:
I don't totally understand you question but if its 4 pages originally then she takes 2 pages every class, then
56-4= 52 (56 pages minus the 4 she already took).
52/2= 26 (52 pages divided by 2 pages per class)
Answer:
Step-by-step explanation:
Notice that we can get from the x-coordinate of A, 1, to the x-coordinte of A', -2, by subtracting 3 from the x-coordinate of A. More formaly:
Similarly, we can get from the y-coordinate of A, 5, to the y-coordinate of A', 3, by subtracting 2 from the y-coordinte of A. More formaly:
Now we now that to get to A' from A, we need to subtract 3 to the x-coordinate of A and subtract 2 to the y-coordinate. Knowing this, we can create the expression to translate any point of the polygon ABCD to create the polygon A'B'C'D':