<p>median was 1300 i thought</p>
<p>r= d/t.. and the distance was 20/ time was .25 and .5 so 20/.75=26.6666667 "math frac" lolol 80/3</p>
<p>sqrt (2) -1 for the the inscribed circle problems</p>
<p>Driving one...</p>
<p>d=rt for first 10 miles
10=(20)t
t=0.5</p>
<p>d=rt for last 10 miles
10=(40)t
t=0.25</p>
<p>d=rt for WHOLE DISTANCE
20=r(0.5+0.25)
r=20/0.75 or 80/3</p>
<p>Man, I better be getting sexual favors from some of the ladies in the house I'm helping out if I ever meet you guys later jk...but seriously.;)</p>
<p>Isn't the inscribed circle [(sqrt)2 - 1 ] / 2?</p>
<p>ooh crap i got .4142 etc.. like just [(sqrt)2 - 1</p>
<p>seriously someone confirm if 30 kids went to college Y and X. Do you add the middle intersection in venn diagram</p>
<p>That's the diameter I believe.</p>
<p>no the diameter is .828, the radius is .4142 which is what the question asked for</p>
<p>I got square root of (2) - 1</p>
<p>For the driving method, Xiggi's formula would have worked fine:</p>
<p>(2<em>a</em>b)/(a+b)</p>
<p>Someone post the solution?</p>
<p>Nope it was sqrt(2)/2 nowitzki!!!</p>
<p>You basically had to draw a diagonal throug hthe picture and realize that the sides of the rectangle formed the sides of a right-angle triangle.</p>
<p>4^2+4^2=c^2
c=4srt(2)</p>
<p>Part of the diagonal in the figure now was the diameter of the two circles which is 4 so you're left with a diameter of sqrt(2) for the litte circle so the radius of it is sqrt(2)/2.</p>
<p>does everyone have the same problems but in different order or all they just all different</p>
<p>evil asian it's SQRT2-1, not SQRT2/2</p>
<p>One fourth of the middle circle was in a square created by radii of one of the bigger circles. The diagonal of the square was sqrt(2), but you had to subtract the radius of the big circle [1] to get the radius of the little circle. </p>
<p>sqrt(2)-1.</p>
<p>Yep...there goes my 800 folks.:(:(</p>
<p>I need a hug lol.</p>
<p>yup, tahts wut i got..</p>
<p>sqrt (2) -1. definitive.</p>
<p>In the first two minutes, I have no clue about the problem.
In the next two minutes, i remember my old trick,and approximate the radius by using my eye measurement.
In the last two minutes, i use 45-45-90 rule to prove my eye-gauge</p>
<p>Is this your first time, evil_asian?</p>