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swin2013
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Why don't I have to factor out the (1+h)^2? on problem 2, page 2?
http://www.math.wisc.edu/~hernande/Teaching/UM/Math115Sec011Winter2011/Quizzes/Quiz3/Solution/Quiz3Sol.pdf
 2 years ago
 2 years ago
swin2013 Group Title
Why don't I have to factor out the (1+h)^2? on problem 2, page 2? http://www.math.wisc.edu/~hernande/Teaching/UM/Math115Sec011Winter2011/Quizzes/Quiz3/Solution/Quiz3Sol.pdf
 2 years ago
 2 years ago

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swin2013 Group TitleBest ResponseYou've already chosen the best response.1
yea but i don't need to square 1+h because why? lol
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
usually i have to square it... that's why. especially for a slope problem.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I don't understand the question. you are asked to find the average velocity = distance/time the distance traveled in the time between 1 and 1+h (with h=0.1) is 4(1.1)^2 +3  ( 4*1^2 +3) 4*1.21 + 3 4 3 4.84 4 0.84 the time interval is 0.1 sec, so velocity is 0.84/0.1 = 8.4 sec
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
in other words, you do square (1+h)
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
but when i squared (1+h), the 4's and 3's cancel. i get 8.04 when the answer should be 8.4 m/s
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
in each case, you find how far the particle got by time step 1. Then find where it is at time step 1+h. The difference between those two distances is how far it moved in h secs.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
but when i squared (1+h), the 4's and 3's cancel. i get 8.04 when the answer should be 8.4 m/s for case (i), with h = 0.1 1+h is 1.1 1.1*1.1= 1.21 4*1.21= 4.84 subtract 4 0.84 divide by 0.1 to get 8.4 you must have made a mistake doing 1.1 * 1.1 ??
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I am not sure how you did your calculation, but you have a "bug" in it, so it would be good to figure out where you are going wrong...
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
\[4(h ^{2}+2h+1) +34(1)^{2}3\] since t =1?
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
and t = 1+h
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes. that looks ok. then 4h^2 +8h +4 +3 4 3 = 4h^2+8h if we divide by h (to get velocity) 4h+8 with h=0.1 this is 4*0.1+8 = 0.4+8= 8.4 perfect!
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
ok that's where i went wrong. i factored out 4h so it would be 4h ( h^2 +2) and canceled out the h's so it would be left with 4(h^2+2)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
if you factor out 4h, it should be 4h(h+2) and 4(h+2) after dividing by h
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
yea that's my mistake... i left the squared. thanks :)
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
i can't believe it was that miniscule
 2 years ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.1
i know! such a tiny mistake could cost me an answer by one decimal place _
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
sometimes it just doesn't work out...good luck!
 2 years ago
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