bzzt. Watch those signs. You meant to say
-8(2t^2-39)= 0
2t^2-39 = 0
2t^2 = 39
t^2 = 39/2
t = √(39/2)
I'm stuck with this problem.
h = -16t^2 + 312
the "t" is seconds after object released and h stands for height. We're suppose to set the equation to 0
so far this is what I have:
-16t^2 + 312 = 0
-8(2t^2+39)= 0
-8=0 can be disqualified but how do I factor out the 2t^2 + 39 = 0 and find the answer? Please help
5 answers
Sorry about the signs. The answer was suppose to be 2 seconds. That's what I'm not understanding. I'm coming out with what you have- I moved over the 2 and that left the t^2 = 39/2 but that doesn't come out to be two seconds which is suppose to be the answer
well, surely you can see that
h(2) = -16(4) + 312 = 248
which is not zero. So, either 2 is wrong, or the equation is wrong.
h(t) = -16t^2 + 64
has h(2) = 0
h(2) = -16(4) + 312 = 248
which is not zero. So, either 2 is wrong, or the equation is wrong.
h(t) = -16t^2 + 64
has h(2) = 0
thank you I'll check back with my teacher.
how do you solve x+y=2?
1 answer