<p>hey maybe im just being stupid but i am totally stumped on #18 on pg. 657</p>
<p>graphically the answer seems right but don't know how to put the math behind it.</p>
<p>1st one ive been stumped on (although i make dumb mistakes). any help would be appreciated.</p>
<p>HOW DO YOU DO IT?</p>
<p>ok</p>
<p>so at t=0 we can represent the function h(t) = c - (d-4t)^2 as
6 = c - d^2</p>
<p>we also know that
106 = c - (d - 10)^2</p>
<p>now let's equate the two equations by making them both equal to c
106 + (d-10)^2 = 6 + d^2</p>
<p>we get d=10. plug it into one of the original equations (6 = c - 10^2) and we get c = 106. now we have</p>
<p>h(t) = 106 - (10-4t)^2 and now we can solve for t = 1. when we plug 1 into t the answer is 70, which i believe is the correct answer.</p>
<p>weird problem. i was trying to plug in c=6. thanks.</p>
<p>Hey, while we're on the subject, can someone explain this one please-- pg. 476, #18.</p>
<p>hmm this might be hard to explain but i'll try.</p>
<p>the easiest way to explain it, but you might not get it w/ this wording, is to use the 4x3x2x1 approach. Since there are 5 cards, there is 4 options at 1st (1 can be matched w/ 2,3,4,5) & then 3 (2 can be matched w/ 3,4,5) & then 2 (3 w/ 4,5) & then 1 (4 w/ 5).</p>
<p>Using this we have 24 combinations. But to find the answer, we have to multiply the 24 by the amount of spaces the shaded square can be in, which is 3. Therefore, 24x3 = 72, which is the answer.</p>
<p>Tell me if that works for you, if not I'll see if I can come up w/ a better explanation. The best way is to kinda chart it out...obviously not for all 72, but it will help give you an idea of the pattern you're looking for.</p>
<p>Hope that helps!</p>
<p>dude this is like some precalculus math and college board said there is only gonna be some algebra 2 stuff
how is this????</p>
<p>I got the right answer, but I looked at it differently. </p>
<p>I figured that if you could put any of the five cards in any space you would have
5x4x3x2x1=120</p>
<p>But you couldn't put one card at either end, so at two places you had</p>
<p>4X3x2x1=24</p>
<p>so you have to take 120 and then take away 24 for one end and 24 for the other end and you end up with 72.</p>
<p>I don't know if this is the right way to do it, or if I just got the right answer by luck.</p>
<p>why doesnt it work if u say, space one has 4 choices, because u cant use the shaded one, then second space has 4 choices beacsue u can use the shaded ont plus the other 3, then 2, then 1 and u get 96. Also for the shoot arros, i just figured since it starts 6 above ground ur force on it only makes it go 100, it does that in like some amount of second, i forget, divide 100 by that number u get how fast it goes per second, i believe u get 40, and since u start above ground answer is 46. i dunno know if its right but i dont see how it wouldnt be.</p>
<p>Hey Islander2588,</p>
<p>You could, feasibly, get the answer to 476 #18 the way you suggest, IF you fill in all the spaces where the gray CANNOT be first. So, first space, 4 possibilities. Last space, 3 possibilities, 2nd space 3 (including gray), 3rd space 2, 4th space 1. 4x3x2x1x3.</p>
<p>I always find it easier to write out the three basic possibilities for the grayed out block first. Place it in space 2, then fill in the other spaces. That give you 4xGRAYx3x2x1=24 possible arrangements IF the block is in space 2. Then move the block to space 3 and do it again. You'll get the same number of possibilities unless the question is really weird, in this case 4x3xGRAYx2x1=24 again. WHen the grayed block is in place 4, same 24 possibilities. Then just add them: 24+24+24.</p>
<p>The problem with your method for P657 Q18 is that the ball doesn't travel at a constant velocity...it's constantly accelerating downward (gravity, ya know). The only reasonable way to solve it is simultaneous equations and substitution as mentioned above. Tough one.</p>
<p>The answer is 72. You have to do exactly what the other person said.
It is like you had five people to arrange in five seats. however, one person could not sit at either end. thus, in the first seat on the left end, there are only four people who could fill that spot. Now that you have used one of the people and the other person can not sit at either end. you next fill the end space on the right. There are only 3 people left who could fill that space. Now you fill the three spaces in the middle. There would only be 2 people left to fill those spaces, but you now pick up the person who could not sit at either end, so there are still 3 people left to fill those 3 spaces. It then becomes 3 factorial in the middle; 3 for the center space on the left, 2 for the center space in the middle and lastly, 1 or the last person for the center space on the right. Thus, permutation is: 4 X 3 X 2 X 1 X 3 = 72</p>