<p>While i was doing some sat math, there was one question that i didnt understand at all.</p>
<p>n(t) = 0.5t² - 20t + k</p>
<p>There was a 100 day period when the number of bees in a certain hive could be modeled by the function n above. in the function, k is a constant and n(t) represents the number of bees on day number t for 0 ≤ t ≤ 99. On what number of day was the number of bees in the hive the same as it was on day number 10?</p>
<p>The correct answer is (B) 30.</p>
<p>Well, im sure i would have had a better chance to find the solution if i understood the question. Can someone please explain me what they mean and how to solve it?</p>
<p>Uhh where is something divided by 2, october4th? If you could explain what I'm missing/what I'm doing wrong it would be great. Thanks!</p>
<p>n(t) = 0.5t^2 - 20t + k
n(10) = 0.5(10)^2 - 20(10) + k
n(10) = 0.5(100) - 200 + k
n(10) = 50 - 200 + k
n(10) = -150 + k
<em>set .5t^2 - 20t equal to whatever you get</em>
0.5t^2 - 20t = -150
Did I do that right/where do I go from here? I tried the quadratic formula on 1/2t^2-20t + 150 and got 20 plus or minus square root of 300, which isn't correct. : (
EDIT: Nevermind, I messed up on the quadratic formula, now I got 30 and 10 :D</p>
<p>Well, I typed the equation into my calculator. k can be anything. I found the y to be 149 when I used find value =10. Then I graphed y =149. Using "intersect" I found x to be 30.
Too lazy to do the algebra.. he.</p>
<p>Sasquatch219, i multiplied by 2 so it would be easier to factor. i ignored k since it is a constant. So k wouldnt really matter unless you are trying to find something in terms of k.</p>
<p>This is a happy parabola. The k doesn’t matter (to get to the answer). Plug in the 10 for t and find that the number of bees in 10 days is -150!. (Doesn’t make much sense - that’s what the k is for). That’s really one half of the parabola. There’s a match on the other half - it can be found by setting -150 equal to .5t^2 - 20t (remember - we’re disregarding the k). From here, you will find that -150 bees (are present?) on the 10th and 30th days.</p>
<p>Here’s a tip for new users: check the dates on a thread before you contribute. You’ve just answered a four-year-old question. Odds are, nansky, the original poster, is well into college by now.</p>
<p>Please don’t resurrect old threads like this one.</p>