<p>^ yeah i second that nigcrunch. it is k≤0</p>
<p>f(4.7) = f(3.7). but f(3.7) still larger than 1… by the same principle f(3.7) = f(2.7) = f(1.7) = f(0.7) (now we have a value such that 0<= x <= 1) = 0.7.</p>
<p>dang itttttt, like 6 skipped and 8 wrongish then lol</p>
<p>is that still above 700???
anyone?</p>
<p>it was
f(X) =
if 0<x<=1 (something=“” like=“” this,=“” then=“” f(x)=“x.” if=“” x=“”>1, then f(x) = f(x-1). something like this.
so you plug in 3.7 into f(x-1) and get f(2.7) so you plug it in again and get f(1.7), plut it again and u get f(.7) which is .7</x<=1></p>
<p>For the .7 one if x is greater than 1, the number keeps going back in and getting one subtracted until it is less than one.</p>
<p>i thought there are two pieces… and if the x value doesnt fit in one piece, you plug it into the 2nd peice. the 2nd piece being x-1, for x>1</p>
<p>and if x was 0<=x<=1 , then thats waht x was. for instance, say they gave you x=.5 then the answer would be .5 because it fits into the 1st piece. if they gave you 3, then it doenst fit in the first piece, so plug it into the 2nd piece, and you get 2. if you keep subtracting every x value >1 than you dont end up with a FUNCTION because then you end up with a black box for the range 0<=x<=1. think about it… for every value of .5, 1.5,2.5,3.5 the way you do it… the answer would always be .5 and for .4, 1.4,2.4,3.4 the answer would always be .4… if thats the case… it doesnt end up being a function</p>
<p>I dont understand the r=cos(theta) question where it wasked you if any of the points where on the line or w.e</p>
<p>can anyone explain that one?</p>
<p>I tried doing a polar graph on my calc but i realized i had no idea what i was doing.</p>
<p>Congruent triangles are triangles that have the same size and shape. This means that the corresponding sides are equal and the corresponding angles are equal. ([Congruent</a> Triangles (with worked solutions and videos)](<a href=“http://www.onlinemathlearning.com/congruent-triangles.html]Congruent”>Congruent Triangles (video lessons, examples and solutions))) Hence non congruent triangles are those that are not equal in size or shape. So the q was asking for how many UNIQUE triangles can be formed. It was a referance to the ambiguous case, does it exist or not. Since no triangles the answer is 0</p>
<p>polar coordinates are in the form (r, thetha) so all u had to do (because r = cos thetha) was find the cosine of the angle and see if it equalled r. Only one that did that was 0.5, pi/3</p>
<p>actually i dont know… watver. i got 3 wrong so far…</p>
<p>ahh okay thank you! I should have reviewed that more.</p>
<p>For the question with the isosceles triangle with sides 5-10-10, you had to use the Law of Cosines right?</p>
<p>Yeah, law of cosines would do the trick, i think the angle was like 29. something. If i remember correctly</p>
<p>Well I divided in half and used Pythagorean for that one</p>
<p>yea it was like 29.3 i think</p>
<p>And yes the angel was 29, either way worked</p>
<p>I used reasoning for that one. the angle had to be less than 90 first of all. so that only left 60 and 29.3. and since the other 2 angles have to be (180-thisangle)/2 60 would mean that (180-60=120/2=60) it would be a 60-60-60 equilateral triangle. and thenit wouldnt be 10-10-5. so it had to be 29.5.</p>
<p>you’re right though law of cosines where C^2=a^2+b^2-2abCos(c) is how you solve it I think. not 100% sure on that.</p>
<p>Now, I used the law of cosines for the last one and got 1 triangle that fit the description and agreed with the triangle inequality theorem. Anyone want to explain why that is wrong? And the k question was so dumb, or at least I thought it was. I put less than(but not equal to) because I thought the graph needed to at least be in the third and fourth quadrant. If k equals zero, none of the graph is in those quadrants.</p>
<p>u cud use law of cosines and do the longer way or you can plit it into two right triangles, use law of sinces, and double the angle.
it was 29</p>
<p>You couldn’t use law of cosines for the second to last one because i believe you need to know two of the angles.</p>