if the distance covered by an object in time is given by s(t)=t^2+5t where s(t) is in metersand t is in seconds, what is the distance covered in the interveral between 1 second and 5 seconds?
8 answers
s(5)-s(1)
A 24 meters
B 30
C 40
D 42
E 44
B 30
C 40
D 42
E 44
so what would be the answer
plug in t=5 and t=1. subtract one from the other. what do you get? you've posted a lot of similar problems. By now you should be getting the hang of them.
30 ahh i get it now thanks
ummm. what did you do?
s(5) = 5^2 + 5*5 = 50
s(1) = 1^1 + 5*1 = 6
s(5)-s(1) = 44, not 30
s(5) = 5^2 + 5*5 = 50
s(1) = 1^1 + 5*1 = 6
s(5)-s(1) = 44, not 30
i did it like this
5^2=25 + 5^1= 5 = 30
i guess i did it wrong
5^2=25 + 5^1= 5 = 30
i guess i did it wrong
that's 5t. not 5^t
5*5 = 25
you did it wrong in two ways. you tried to do 5^2 + 1^2 but did 5^2 + 5^1 instead.
You did not include both terms in your evaluations.
*sigh* review your algebra, if you want to do well in calculus.
5*5 = 25
you did it wrong in two ways. you tried to do 5^2 + 1^2 but did 5^2 + 5^1 instead.
You did not include both terms in your evaluations.
*sigh* review your algebra, if you want to do well in calculus.