<p>I put I and III also!</p>
<p>We all win! Lets have a online cookie bash.</p>
<p>I thought it was hard, they had a crap load of variable problems</p>
<p>it was easy but i didnt ****ing finish! 9 blank and i knew every answer!</p>
<p>it was incredibly easy compared to the physics</p>
<p>ditto to crossover, i got wrong i think mixing up x and y intercept</p>
<p>Already a discussion. Look below.</p>
<p>the log question that was:
log2(x^2 + 12 x) = 6
becomes
x^2 + 12 x =2^6 = 64
x^2 + 12 x - 64 = 0
(x-4)(x+16) = 0
x = 4, x=-16</p>
<p>-16 was the answer (I thing it was answer choice E?)</p>
<p>-Aaron</p>
<p>I left two blank and got at least one wrong, but that will mean I did very well.</p>
<p>-the log question that was:
-log2(x^2 + 12 x) = 6
-becomes
-x^2 + 12 x =2^6 = 64
-x^2 + 12 x - 64 = 0
-(x-4)(x+16) = 0
-x = 4, x=-16</p>
<p>--16 was the answer (I thing it was answer choice E?)</p>
<p>--Aaron</p>
<p>was it y=6 or x=6 i forget which, but i think i did it so that x=6 and came out to 6.8 which was an answer</p>
<p>it was y =6 answer -16</p>
<p>that sucks</p>
<p>the last question, what is the way to solve that with out intergrals?</p>
<p>i did it after the test and it comes to a little more then 8 but i coudlnt set up my intergral intime (had like 3 mins on it) ont he test</p>
<p>last question: with the A+B+C in the digits of the interger</p>
<p>sum of other digits = 22
divisilbe by 3 3 times, needs to add to 3^3 =27 so A+B+C = 5
answer a</p>
<p>that waste the last one, the one witht he area of the trapzoid formed by 4-^2 and a lien that turn out to be y=2.43</p>
<p>the last question read something like:</p>
<p>you have 4,ABC,239,040. what is A+B+C equal if the number is divisible by 3 consecutive multiples of 3.</p>
<p>It doesn't specify which three, so I chose 3 6 and 9.</p>
<p>first thing I did was to sum up the digits which gives you 22+A+B+C.
remember in like fifth grade when you were taught division tricks? well those come into play now:</p>
<p>if you sum up the digits and they are divisible by 3, then 3 can go into that number. that gives us A+B+C = 2, 5, or 7 (the sums would be 24, 27, 30 respectively.)</p>
<p>next we check for 6... if 3 goes into it (same conditions as above apply here then) and 2 goes into it, it is divisible by 6. it is even. therefore automatically divisible by six.</p>
<p>We come to the final multiple. 9. the rule for nine is the same as the one for 3: if the sum of the digits is divisible by 9, then 9 goes into that number. the only number in 24,27,and 30 that 9 goes into is 27.</p>
<p>Thus A+B+C = 5</p>
<p>To check this, choose any A B and C that add up to 5 and divide 4,ABC,239,040 by 3, 6, and then 9. it should give you whole number results.</p>
<h2>-Aaron</h2>
<p>looks like mikenthemaddog66 beat me to it</p>
<p>the trapizoid/parabola one I got y=2.4375 and the area {[(2.5+4)/2][2.4375]} equals 7.92</p>
<p>-Aaron</p>
<p>That is correct.</p>
<p>You had to use the knowledge that the x coordinate was half of the given length between the two points - 2.5/2 (I think it was 2.5) and then use that to find the y value to find the height of the trapezoid.</p>
<p>i still dont get how you did it when the two sides are curves with out intergrals</p>
<p>and calculus tells me it was 8.03 in chance which was a choice</p>
<p>Does anyone remember the function so I can show how I did it?</p>
<p>And, hugh, if you used calculus, you found the wrong value. They didn't want to know the area under the curve, they wanted to know the area of the trapezoid only.</p>
<p>For the trapezoid/parabola question, I just used calculus to find the area from -2 to 2 and subtract the area of the upper portion which was given.</p>