mathmathj28

In this problem, the value of the product wxyz is not given, so I don't think I can solve it in the same way. I'm also more interested in how to solve the problem than just the answers. Thanks for linking the other question and for your help - I enjoyed reading heureka's solution!

Thank you so much for your help!!!

Thanks so much for your help!!!

Could you please explain how you found those sums? I don't quite understand.

After working on this problem for a while, I found that the sum of the first row is \(\frac{1}{1-a}\) because the formula for the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), where r=the common ratio. Similarly,  I found that the sum of the second row is \(\frac{b}{1-a}\), and the sum of the third row is \(\frac{b^2}{1-a}\). This led me to believe that the sum of each row could be multiplied by b to get the sum of the next row, and therefore, the sums of the rows form a geometric series. I then found the sum of that series in the same way. Since the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), I found that the sum of all the squares in the grid is \(\frac{\frac{1}{1-a}}{1-b}=\frac{1}{(1-a)(1-b)}\). I haven't worked on part b yet, but I assume the sum can be found the same way.

The difference in our solutions has me wondering if I made a mistake, and if so, where I made it. I would really appreciate if someone could help me out with this!

Thank you for the link!! I solved it the same way as Melody, but I am also struggling to find a way to prove it, so if possible I would like some help with that. 

Thank you so much for the explanation!!!

I can't really tell whether one pair of faces is at an angle. If so, it is a truncated pyramid, like Electriv Pavlov said. If not, it's a trapezoidal prism. Hope this helps!

Thanks for the help!! I believe a-2b should equal 7, though, not -7.

Thanks!!!

Thanks for the help!!

Thank you so much!!

If the mean of his scores is 57 and there are 5 judges, then his total score was 57*5=285. If each judge gives out a score greater than 50, then the lowest score a judge can give is 51. To find the highest amount of points he could have earned from one judge, we can assume that the other four give him the lowest possible score, 51. x+51*4=285 simplifies to x=81, so the highest score he could have earned from one judge is 81.

Thank you so much for the help Alan!! I didn't think of doing that before but I understand why that is helpful now.

If my work is correct, the sum of the coefficients of the terms in g(x) that have even degree is 21,556,128.5

\(g(1)=f(1)^{16}=(-\frac{3}{2}+\frac{1}{2}+4)^{16}=3^{16}=43,046,721 \)

\(g(-1)=f(-1)^{16}=(-\frac{3}{2}-\frac{1}{2}+4)^{16}=2^{16}=65,536\)

\(\frac{g(1)+g(-1)}{2}=\frac{43,046,721+65,536}{2}=\frac{43,112,257}{2}=21,556,128.5\)

Thanks again for helping me solve this!

Could you please explain how you got there? Thanks

Could you please explain how you solved it? I got a different answer. Thanks

Thanks so much!!