<p>In the figure above, Triangle ABE is equilateral and Triangle EBD and Triangle DBC are isosceles. What is the value of x ?</p>
<p>I do not get this question at all...an explanation would be rad.</p>
<p>x = 15 ???</p>
<p>Yes, sorry, I forgot to post the answer.</p>
<p>well,triangle ABE is equilateral so angle AEB = 60 ==> therefore angle DEB = 180-60 = 120.
EBD is isosceles so angle BDE = EBD = (180-120)/2 = 30
If EDB = 30 degrees ==>CDB = 180-30 = 150 ,then again ,x = (180-150)/2 = 15</p>
<p>Ahhhhhh ok thanks a lot, now I understand. Just wasn’t clicking :D</p>
<p>Another Question:</p>
<p>How is (2^m)(2^3m) when m = 2 equal to 256?</p>
<p>You would multiply the 2’s, giving you 4, and then multiply the m’s, giving you m+3m = 4m… </p>
<p>4^4m = 4^4(2) = 4^8 = 65536… I’m confused on that</p>
<p>You don’t multiply the 2’s, you just add the exponents.</p>
<p>(2^m)(2^3m) = 2^4m = 2^8 = 256</p>
<p>Hahaha I’m ■■■■■■■■…it’s been a while since algebra II, I’m a bit blurry.</p>
<p>Don’t worry, I do things like that all the time :)</p>
<p>Hahaha. This thread totally made my day! I do this ALL the time!</p>
<p>Lol, another question which I’m sure is insanely easy but I’m missing something…</p>
<p>The circumference, in inches, of a certain circle is C, where |C-20.5| Less than or equal to 0.5 . Which of the following is a possible area of the circle, in square inches?</p>
<p>The answer is 100/pi</p>
<p>Annnnd this one: <a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
<p>In the figure above, l || m and each circle is tangent to the other two circles at the labeled points. If each circle has diameter 8 and is tangent to one of the two lines, what is the distance, h, between l and m ? The answer is 4 root 3 + 8</p>
<p>Lol i am not sure I got this question but
to solve this kind of questions ,you need to work backwards</p>
<p>I try each answer starting with D and see what works
so ,lets try your answer
A = p(r)^2
100/p = p(r)^2
100=p^r^2
pr=10
r = 10/p</p>
<p>C = 2pr
C = 2p(10/p)
C=20
See if 20 works</p>
<p>/C-20.5/ is equal or less than 0.5
/20-20.5/ is equal or less than 0.5
/-0.5/ is equal or less than 0.5
0.5 = 0.5
So it works
100/pi is a possible solution.This takes time but I cant think of a faster way.I do think ,however ,that this is not so easy problem and should be one of the last in the section</p>
<p><a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
abot this,
there is not easy way to explain it but each center of the three triangles is a vertex of equal triangle with sides of 8 .
THe altitude of this triangle is sqrt 4 root 3 .The first vertex of the altitude is the center of the upper circle and the second is the point where the two circles is point .If you continue that line to line l and line m ,you will see that it is 4 root 3 + twice the radius ==>4 root 3 + 8
I realize I didnt explain it well because I dont know many of the math terms in English ,but I hope I helped :")</p>
<p>Thanks for the explanations, I get those ones now 
One more question</p>
<p><a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
<p>How the heck do I find the length of RS? The larger circle has radius of 6 and the smaller one has radius 1. Thanks</p>
<p>There are two ways to approach it. If you extend PQ to become the diameter, then easy Power of a point application shows that the answer is sqrt(13) - 1. </p>
<p>Approach 2: RQ is tangent to the circle, and as a result is perpendicular to RP. This makes a right trinagle. RP^2 + RQ^2 = PQ^2
RP = 6
PQ = 7</p>
<p>RQ = sqrt(13)</p>
<p>RQ = RS + SQ
sqrt(13) = RS + 1
RS + sqrt(13) - 1</p>
<p>Ohhhh ok, I didn’t really get what you were saying with Method 1 but Method 2 makes a lot of sense. Thanks</p>