question
q.8  2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. find the number of rows and the number of plants in each row.
337
The next shapes in the ‘growing’ pattern are as follows:


![](https://mathsgee.com/?qa=blob&qa_blobid=5951663371716016349) finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). ok, in that case replace these two lines
 last = lastrow(destsh)
 shlast = lastrow(sh)

with
 last = destsh.range("a" & rows.count).end(xlup).row
 shlast = sh.range("a" & rows.count).end(xlup).row ![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/vvnzofse6383734697724592802%20png)


let the length of path be a m


total area ![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/ceaxujzx7923478049123545928%20png)


area of lawn = total area â€“ area of
path

![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/1800204/2012_12_04_21_09_56/mathmlequation2002748073868866733%20png)


area of lawn


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/dwqusyms3311632565229211681%20png) i would do it like this:

https://www.advanced-excel.com/columns-to-rows.html answer: 15
solution: there are six straight lines `l\_(1),l\_(2),l\_(3),l\_(4),l\_(5),l\_(6)` in a plane.   
 since no two lines are parallel and no three of which pass through the same point, we have maximum points of intersection.   
 line `l\_(1)` intersect five lines, so there will be five points of intersection.   
 similarly, for each of the lines `l\_(2),l\_(3),l\_(4),l\_(5),l\_(6)`, there will be five points of intersection.   
 but there will be double counting .   
 so, total number of points of intersection is `(5xx6)/(2)=15` Hi Djhenry,
  

  
When you right-click a sheets tab, you can select all sheets. When you have done that, you can insert a row and enter all the information and all sheets wil contain that information.
  

  
Best regards,
  
Trowa here's my guess. you create x1pts and x2pts without specifying the number of elements you want. when you do that, you get a vector with length 100. see this comment from the documentation page:
y = linspace(x1,x2) returns a row vector of 100 evenly spaced points between x1 and x2.
therefore, x1grid and x2grid and 100x100.
i have no idea what the size of posteriorrange is, but max(posteriorregion,[],2) must not be a 10000 x 1 (size(x1grid,1) = size(x1grid,2) = 100). g = findgroups(yourtable.name);
output = splitapply(@(varargin) {table(varargin{:}, 'variablenames', yourtable.properties.variablenames)}, yourtable, g);
this should give a cell array of tables. I have figured out this issue and so for anyone else with the same issue.

In Designer for your XDP make sure you have a subform as the entry for the table row, and make this subform "Repeat Subform for Each Data Item" I found the subform could not be within another subform for this option to be visible.

So after this you can assign the row to the subform and the elements to the elements within the subform and hopefully it'll work.




[View solution in original post](/t5/adobe-experience-manager-forms/dynamic-rows-in-a-table-issue-also-relates-to-having-elements/m-p/204725#M6246)