<p>it was greater than 90, cuz when i solved it throught 30-60-90 by plugging in the x, 2x, and xsqt3, it wasn't possible that the angle was strictly between 60 and 90</p>
<p>3,4,6 was a 30-60-90 triangle?</p>
<p>Thanks...I'm hoping....that's what's going to happen !_!</p>
<p>about the math, if you know Laws of Cosines...that's how 3,4,6 could be done</p>
<p>c^2 = a^2 + b^2 - 2(a)(b)*cos(angle c)</p>
<p>So, if you set a = 3, b = 4, c = 6, you can find angle c > 90</p>
<p>Here are some answers I remember putting down for the grid-ins:</p>
<p>15, 4, 115, 1600, 3/4, 10</p>
<p>Anyone remember other numbers?</p>
<p>^ 9, 18 (10char)</p>
<p>.8, .167 (it was .166666) 9</p>
<p>What was "9" an answer to?</p>
<p>Oh, and also for the x < 2 < 3x or something, I had .9 but I think it was a range of answers.</p>
<p>oooooooo! i do!! all of them except 3/4...i didn't finish the grid ins though</p>
<p>My God I did so badly in the maths.... this is just embarrassing.</p>
<p>What was .8?</p>
<p>admanrich- x<2<3x</p>
<p>OOO ok good i did .75</p>
<p>1) 15
2) 4
3) 1/6
4) .9 (this could have also been .66 to .9)
5) 115
6) 18
7) Can't remember
8) 1600
9) 3/4
10) 10</p>
<p>What was the #7 grid-in? Was that the 9? I can't seem to remember that problem.</p>
<p>as long as it was between .7-.9, it's right</p>
<p>Yup those look right-- I missed 10 but w/e.</p>
<p>Yeah I think #7 was 9--it was the one "a^x/a^y = a^-6-- I remember it was second row towards the left.</p>
<p>for my math grid in .. the answer were </p>
<ol>
<li>1 or 2</li>
<li>15</li>
<li>Forgot</li>
<li>Forgot</li>
<li>Forgot</li>
<li>2000</li>
<li>9 </li>
<li>2.5</li>
<li>3/4</li>
<li>10</li>
</ol>
<p>was the second problem with roman numerals II or II and III. I cant remember exactly what the problem was. I put II and III. Does anyone know what the problem was exactly</p>
<p>I can't remember the problem either. This is frustrating :P</p>
<p>The two-gallon problem was tricky. Here's my solution to it:</p>
<p>(Name) has a scooter (motorbike?) that runs m miles per 1 gallon. Her scooter can hold 2 gallons of gas and gas costs $1.25 per gallon. If she pays d dollars to fill her tank, how much gas was there in the scooter/motorbike initially?</p>
<p>The hardest thing about this problem was realizing which variables are relevant.</p>
<p>Using an equation involving cost, I came to my answer.</p>
<p>$d + $1.25g = $2.5, where g is the gallons needed to fill the tank.</p>
<p>d + 1.25g = 2.5
2.5 - d = 1.25g
g = (2.5-d)/1.25
g = 2 - d/1.25</p>
<p>Haha Barb and her motorbike-- I hated that problem it was so unrealistic these days with the gas prices haha</p>