October 25, 2008 ACT Discussion- Math

<p>twiz91 - i think with the trapezoid the top was 4 and the bottom was 14, so what i did was subtract 2 from 7 to get one side and then i think 12 was the other, so i used pythag. and i think i got 13. then 13+13+4+14 = 44. that’s what i did, then again i was rushed</p>

<p>Oh wow that would’ve been a much easier way to solve it rofl. I did it the LONG pythag way haha.</p>

<p>and why don’t i remember this boat one at all, can someone explain to me what we had to do, im going crazy trying to remember.</p>

<p>I think it was 44 for the perimeter of the trapezoid. The base was 14, the top was 4, and I made triangles with a height of 12 and base of 5. The sides were 13.</p>

<p>Boat problem…</p>

<p>You could use the law of cosines like they said which was you would have had to make a triangle for with the horizontal line and the one at 40 degrees. Then you would have had to make the angle you know inside the triangle, the 140 (180-40) angle A and used the law of cosines that they gave you to get the side opposite to it which was the length from the beginning to the end of the trip. You knew the other 2 sides cause they told you the lengths so when you plugged it all into the equation they gave you you would have gotten something like 51.7… Or if you know about vectors you could do vector addition which I think is way easier. If you break up the vector at an angle to its horizontal and vertical components you could just add the two horizontals (since the other was already horizontal) and thats one side off the right triangle then the other side is the vertical component and you could do pythagorean theorem and get the same answer.</p>

<p>and why don’t i remember this boat one at all, can someone explain to me what we had to do, im going crazy trying to remember.</p>

<p>yeah lol i don’t remember a boat either.</p>

<p>oh yes i remember. thank you. i believe that took me the longest</p>

<p>The fan one was sooo confusing! I put 60 degrees because it was in between 45 and 72 roflcopter.</p>

<p>The fan one was sooo confusing! I put 60 degrees because it was in between 45 and 72 roflcopter.</p>

<p>Each fan part is equal to 360(1/5) = 72. If one fan part moves exactly one fan part, it moves 72 degrees.</p>

<p>boat was when they had two distances making an obtuse triangle without the top side. i believe the distances were 25 and 30, and they asked for the straight line distance from start to stop ,and they addressed the law of cosines. </p>

<p>that was one of the ones that i thought would take a bit more time so i skipped it initially. when i came back to it i eliminated like a d and e just because the measures seemed two extreme for it to be the distance. and i guessed 52 woooo.</p>

<p>another thing, are the figures supplied drawn to scale?</p>

<p>what about that problem with i ?</p>

<p>what about that problem with i ?</p>

<p>The answer was something like (blah)(blah)(x^2 + 1)</p>

<p>yes llpitch i remember that it confused me for a second and okay i guessed that too on the boat one.</p>

<p>problem 60, asked which equation has the solutions, 1/2,3/4, i, i^2</p>

<p>It was choice A</p>

<p>(2x-1)(4x-3)(X^2+1)</p>

<p>^ That’s what I got too.</p>

<p>I think the figures are at least close to scale unless they say not to scale but I might be wrong. </p>

<p>And for the i one, if you have roots you know the polynomial has to be some integer multiple of x-each root multiplied together, and to get i you have to take the sqrrt-1</p>

<p>(x-1/2)(x-3/4)(x^2+1)</p>

<p>Since x^2 + 1 = 0 was the only way you’d be taking the sqrrt of -1 which has the same roots as
2(x-1/2)4(x-3/4)(x^2+1)
Since multiplying by a number doesn’t change the roots
which simplifies to one of the answers
(2x-1)(4x-3)(x^2+1)</p>

<p>(didnt notice someone else just did this one above me LOL)</p>

<p>have to agree with baylay</p>

<p>sweet thats what i got! thanks everyone :)</p>

<p>(2x-1)(4x-3)(x^2+1)</p>

<p>Yup that was it. I forgot the first two terms lol.</p>

<p>Did anyone choose the answer 5x+6y=30(y=-5x/6+5) for the perpendicular line? prob 59</p>

<p>the equation of the line they wanted it perpendicular to was 6x-5y=30</p>

<p>^ Thats what i got.</p>