<p>the -f(x) was -3.</p>
<p>101 percent sure.</p>
<p>i got 10pi</p>
<p>-f(x) question? What’s that? For the directly proportional problem, was the answer 10?</p>
<p>pat can you explain that one? i guessed correctly…</p>
<p>I think it was:</p>
<p>A) 16
B) 20
C) 24
D) 16 + 2root3
E) 16 + 4root2</p>
<p>Or something…</p>
<p>
</p>
<p>10 pi was the answer.</p>
<p>the length of AB was 20, there were 5 sections.
each diameter was 4; the circumference was 4pi</p>
<p>since there are 2.5 circles, its 4pi x 2.5 or 10.</p>
<p>another way was to divide 4pi by 2 because there are semicircles.</p>
<p>so you get 2pi for each semicircle segment. 2x5 = 10pi</p>
<p>akitorin, isn’t that the trapezoid one not rectangle one?</p>
<p>patr,which was that? x^3+x one?</p>
<p>every team had to have the same amount of people</p>
<p>i think 24 might have been an answer choice…i think it was E…but im not too sure</p>
<p>
the equation was x^2+kx-10</p>
<p>look at -10. what are the factors?
5 and -2 or -5 and 2</p>
<p>(x+5)(x-2)
x^2-3x-10</p>
<p>k is -3</p>
<p>(x-5)(x+2)
x^2-3x-10</p>
<p>k is still -3 :D</p>
<p>actually that might be a different problem.</p>
<p>i think the question was like:</p>
<p>what would make both equations equal if x=3 and -f(x)=f(-x)</p>
<p>and hen</p>
<p>the answer
was</p>
<p>x3+x…</p>
<p>ooh,i see that one , yea that was easy</p>
<p>Questions: </p>
<p>Does anyone remember the answer choices for the rectangle that was 8x3 in dimensions?<br>
For trapezoid one, what were the answer choices?</p>
<p>yes pat, but there were no x=3 lol…</p>
<p>-(x^3+x)= -x^3-x
(-x)^3+(-x)= -x^3-x</p>
<p>I thank Math II for this LOL</p>
<p>another way for x^2+kx-10</p>
<p>25 + 5k - 10 = 0 then solve for k</p>
<p>what would -5 or -6 be?</p>
<p>^ i think that was the easiest way to solve it</p>
<p>Thiscouldbeheavn, I did it that way too.</p>