<p>for the question with sum > 5 (answers are like .9, .8, etc.) does anybody know if the range was 0-9 or 1-10? it really makes a difference in the answer</p>
<p>range was 1 to 10 inclusive</p>
<p>For: f(2x)=3x^2, find f(g(x)): 3(2(g(x))^2</p>
<p>Was there an answer choice that said 3(g(x))^2 ??</p>
<p>And why is it 3(2(g(x))^2 if it said f(2x)=3x^2 for any x, doesn’t that mean it doesn’t matter if it’s 2x or x in the f function? Ah I feel confused.</p>
<p>@AforAmbition,
set g(x)=x so f(2x) is f(2(g(x))=3(2(g(x)))^2</p>
<p>@chloe, kk then i think the answer would be .9. </p>
<p>there are 10 possibilities for getting a sum less than or equal to 5:
1+1, 1+2, 1+3, 1+4
2+1, 2+2, 2+3
3+1, 3+2
4+1</p>
<p>100(total combos) - 10 = 90, 90/100 = .9</p>
<p>Dang it… I was going to put .9 but left it blank. -_-</p>
<p>Okay so I left 4 blank and got 4 wrong… gg</p>
<p>Still trying to figure out the line segment by end points…</p>
<p>(0, f(0)) and (pi/2, g(pi/2))
f(t)=2sint, g(t)=cos(t/2)?</p>
<p>(0, f(0)) = (0, 2sin(0)) = (0, 0)
(pi/2, g(pi/2)) = (pi/2, cos((pi/2)/2)) = (pi/2, (2^(1/2)/2)
((pi/2-0)^2+((2^(1/2)/2)-0)^2)^(1/2)= 1.7226 and I’m pretty sure that wasn’t an answer choice.</p>
<p>shouldn’t f(2x)=3x^2, find f(g(x))= (3(g(x)^2))/4
because f(x) would be (3/4)x^2 so that when you plug 2x in you get 3x^2</p>
<p>Also for the period one I put it is not periodic (can piecewise functions be periodic?)… and the P question I got P is Odd.</p>
<p>Tomaras is right, f(x)= (3/4)(x^2) so therefore f(g(x))= (3/4)(g(x)^2) = 3(g(x)^2)/4</p>
<p>f(2x)=3x^2
(3x)(x)=3x^2
(3/2)(2x)(1/2)(2x) =3x^2
f(x)=(3/2)(x)(1/2)(x)
f(x)= (3/4)(x^2)</p>
<p>f(g(x))= (3(g(x)^2))/4</p>
<p>f(2x) = 3x^2 = 3 <em>(1/4 *4)</em>x^2=(3<em>1/4)</em>(4<em>x^2)= 3/4</em>(2x)^2
If f(2x) = 3/4<em>(2x)^2, then f(x)=3/4</em>x^2
by replacing g(x) for x in f(x)=3/4<em>x^2, you get f(g(x)) = 3/4</em>(g(x)^2</p>
<p>3x-4=0, 3x-4 also equals what: 17</p>
<p>wth? i just got 4/3?</p>
<p>@chloeee: Thanks for the compilation!
I’m pretty sure the period was 2. Any function can be defined as periodic, whether or not it’s continuous. I believe that the question stated even numbers = -1 and odd = 1, with anything else = 0. Wouldn’t this make the function continuous (looking kind of like the lines on a heartbeat monitor)?</p>
<p>NspiredOne, so did I, I hope they were talking about another problem :/</p>
<p>@NspiredOne: The question had two roots: (not sure what this one was)(3x-4)=0. So if you plugged in the other x into 3x-4, you’d get 17.</p>
<p>Edit: I think the equation was (x-7)(3x-4)=0. So x=7 as well and if you plug 7 into 3x-4, you get 17.</p>
<p>(x-7)(3x-4)=0
x has to be either 7 or 3/4 because of the zero product property…
so 3(4/3)-3=0
and 3(7)-4=17</p>
<p>@adelice That would cause point discontinuity at every integer. However it may still be able to be periodic? I’m not sure</p>
<p>■■■ 5 skiped 4 wrong… what is that?</p>
<p>Don’t you think the first 30 questions on this test were a breeze…and I was well on my way to finishing before time!! But believe it or not…the last 15 qns were just ah! I ran out of time on number 49!</p>
<p>If i skipped two and missed at least two is it still at all possible to get an 800? Like i said before, I think they will curve this test more than usual…</p>