I need an expression that equals 113079 [closed]

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My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.







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closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – Xander Henderson, Jendrik Stelzner, Tyrone
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
    – John Dvorak
    Aug 9 at 15:39






  • 5




    Did he have a favorite kind of math? Geometry, calculus, etc.?
    – Clayton
    Aug 9 at 15:40










  • Prime factorization: $3 times 37693$.
    – David G. Stork
    Aug 9 at 15:59














up vote
1
down vote

favorite












My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.







share|cite|improve this question














closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – Xander Henderson, Jendrik Stelzner, Tyrone
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
    – John Dvorak
    Aug 9 at 15:39






  • 5




    Did he have a favorite kind of math? Geometry, calculus, etc.?
    – Clayton
    Aug 9 at 15:40










  • Prime factorization: $3 times 37693$.
    – David G. Stork
    Aug 9 at 15:59












up vote
1
down vote

favorite









up vote
1
down vote

favorite











My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.







share|cite|improve this question














My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 9 at 16:23









amWhy

189k25219431




189k25219431










asked Aug 9 at 15:37









JenaJ

171




171




closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – Xander Henderson, Jendrik Stelzner, Tyrone
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – Xander Henderson, Jendrik Stelzner, Tyrone
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
    – John Dvorak
    Aug 9 at 15:39






  • 5




    Did he have a favorite kind of math? Geometry, calculus, etc.?
    – Clayton
    Aug 9 at 15:40










  • Prime factorization: $3 times 37693$.
    – David G. Stork
    Aug 9 at 15:59












  • 1




    113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
    – John Dvorak
    Aug 9 at 15:39






  • 5




    Did he have a favorite kind of math? Geometry, calculus, etc.?
    – Clayton
    Aug 9 at 15:40










  • Prime factorization: $3 times 37693$.
    – David G. Stork
    Aug 9 at 15:59







1




1




113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
– John Dvorak
Aug 9 at 15:39




113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
– John Dvorak
Aug 9 at 15:39




5




5




Did he have a favorite kind of math? Geometry, calculus, etc.?
– Clayton
Aug 9 at 15:40




Did he have a favorite kind of math? Geometry, calculus, etc.?
– Clayton
Aug 9 at 15:40












Prime factorization: $3 times 37693$.
– David G. Stork
Aug 9 at 15:59




Prime factorization: $3 times 37693$.
– David G. Stork
Aug 9 at 15:59










3 Answers
3






active

oldest

votes

















up vote
2
down vote













Here are some ideas that make vague mathematical sense!



$$sum_j=0^infty w_j=113079$$



$$j^3+w^3+J^3=113079$$



$$J(w(j))=113079$$



$$J^11+W^30=J^79$$



$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$






share|cite|improve this answer



























    up vote
    1
    down vote













    There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.



    Unfortunately, the prime factorisation is rather boring:



    $3 times 37693$



    EDIT:



    However, you could do



    $J.W.J. = 2 times 7 times 41 times 197 + 1 $



    or



    $ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $



    Finally, there's also stuff more like this:



    $$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$



    which looks a bit more "equationy".






    share|cite|improve this answer






















    • I really like this one that you call "equationy". Thank you.
      – JenaJ
      Aug 10 at 18:35










    • You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
      – Matt
      Aug 10 at 21:00










    • It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
      – JenaJ
      Aug 12 at 22:00











    • When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
      – Matt
      Aug 12 at 22:23










    • I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
      – JenaJ
      Aug 13 at 23:51

















    up vote
    1
    down vote













    This number is the sum of three cubes, i.e.,
    $$
    113079=J^3+W^3+J'^3,
    $$
    with (large) integers $J,W,J'$.



    Reference: This MO-question.






    share|cite|improve this answer






















    • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
      – Brahadeesh
      Aug 9 at 20:10










    • @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
      – Dietrich Burde
      Aug 10 at 8:51











    • Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
      – Dietrich Burde
      Aug 10 at 9:10











    • Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
      – Brahadeesh
      Aug 10 at 10:07










    • Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
      – JenaJ
      Aug 10 at 18:33

















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    Here are some ideas that make vague mathematical sense!



    $$sum_j=0^infty w_j=113079$$



    $$j^3+w^3+J^3=113079$$



    $$J(w(j))=113079$$



    $$J^11+W^30=J^79$$



    $$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$






    share|cite|improve this answer
























      up vote
      2
      down vote













      Here are some ideas that make vague mathematical sense!



      $$sum_j=0^infty w_j=113079$$



      $$j^3+w^3+J^3=113079$$



      $$J(w(j))=113079$$



      $$J^11+W^30=J^79$$



      $$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        Here are some ideas that make vague mathematical sense!



        $$sum_j=0^infty w_j=113079$$



        $$j^3+w^3+J^3=113079$$



        $$J(w(j))=113079$$



        $$J^11+W^30=J^79$$



        $$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$






        share|cite|improve this answer












        Here are some ideas that make vague mathematical sense!



        $$sum_j=0^infty w_j=113079$$



        $$j^3+w^3+J^3=113079$$



        $$J(w(j))=113079$$



        $$J^11+W^30=J^79$$



        $$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 9 at 16:11









        Malkin

        1,482523




        1,482523




















            up vote
            1
            down vote













            There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.



            Unfortunately, the prime factorisation is rather boring:



            $3 times 37693$



            EDIT:



            However, you could do



            $J.W.J. = 2 times 7 times 41 times 197 + 1 $



            or



            $ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $



            Finally, there's also stuff more like this:



            $$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$



            which looks a bit more "equationy".






            share|cite|improve this answer






















            • I really like this one that you call "equationy". Thank you.
              – JenaJ
              Aug 10 at 18:35










            • You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
              – Matt
              Aug 10 at 21:00










            • It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
              – JenaJ
              Aug 12 at 22:00











            • When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
              – Matt
              Aug 12 at 22:23










            • I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
              – JenaJ
              Aug 13 at 23:51














            up vote
            1
            down vote













            There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.



            Unfortunately, the prime factorisation is rather boring:



            $3 times 37693$



            EDIT:



            However, you could do



            $J.W.J. = 2 times 7 times 41 times 197 + 1 $



            or



            $ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $



            Finally, there's also stuff more like this:



            $$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$



            which looks a bit more "equationy".






            share|cite|improve this answer






















            • I really like this one that you call "equationy". Thank you.
              – JenaJ
              Aug 10 at 18:35










            • You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
              – Matt
              Aug 10 at 21:00










            • It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
              – JenaJ
              Aug 12 at 22:00











            • When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
              – Matt
              Aug 12 at 22:23










            • I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
              – JenaJ
              Aug 13 at 23:51












            up vote
            1
            down vote










            up vote
            1
            down vote









            There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.



            Unfortunately, the prime factorisation is rather boring:



            $3 times 37693$



            EDIT:



            However, you could do



            $J.W.J. = 2 times 7 times 41 times 197 + 1 $



            or



            $ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $



            Finally, there's also stuff more like this:



            $$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$



            which looks a bit more "equationy".






            share|cite|improve this answer














            There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.



            Unfortunately, the prime factorisation is rather boring:



            $3 times 37693$



            EDIT:



            However, you could do



            $J.W.J. = 2 times 7 times 41 times 197 + 1 $



            or



            $ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $



            Finally, there's also stuff more like this:



            $$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$



            which looks a bit more "equationy".







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 9 at 16:17

























            answered Aug 9 at 15:42









            Matt

            1,972617




            1,972617











            • I really like this one that you call "equationy". Thank you.
              – JenaJ
              Aug 10 at 18:35










            • You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
              – Matt
              Aug 10 at 21:00










            • It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
              – JenaJ
              Aug 12 at 22:00











            • When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
              – Matt
              Aug 12 at 22:23










            • I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
              – JenaJ
              Aug 13 at 23:51
















            • I really like this one that you call "equationy". Thank you.
              – JenaJ
              Aug 10 at 18:35










            • You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
              – Matt
              Aug 10 at 21:00










            • It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
              – JenaJ
              Aug 12 at 22:00











            • When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
              – Matt
              Aug 12 at 22:23










            • I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
              – JenaJ
              Aug 13 at 23:51















            I really like this one that you call "equationy". Thank you.
            – JenaJ
            Aug 10 at 18:35




            I really like this one that you call "equationy". Thank you.
            – JenaJ
            Aug 10 at 18:35












            You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
            – Matt
            Aug 10 at 21:00




            You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
            – Matt
            Aug 10 at 21:00












            It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
            – JenaJ
            Aug 12 at 22:00





            It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
            – JenaJ
            Aug 12 at 22:00













            When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
            – Matt
            Aug 12 at 22:23




            When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
            – Matt
            Aug 12 at 22:23












            I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
            – JenaJ
            Aug 13 at 23:51




            I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
            – JenaJ
            Aug 13 at 23:51










            up vote
            1
            down vote













            This number is the sum of three cubes, i.e.,
            $$
            113079=J^3+W^3+J'^3,
            $$
            with (large) integers $J,W,J'$.



            Reference: This MO-question.






            share|cite|improve this answer






















            • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
              – Brahadeesh
              Aug 9 at 20:10










            • @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
              – Dietrich Burde
              Aug 10 at 8:51











            • Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
              – Dietrich Burde
              Aug 10 at 9:10











            • Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
              – Brahadeesh
              Aug 10 at 10:07










            • Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
              – JenaJ
              Aug 10 at 18:33














            up vote
            1
            down vote













            This number is the sum of three cubes, i.e.,
            $$
            113079=J^3+W^3+J'^3,
            $$
            with (large) integers $J,W,J'$.



            Reference: This MO-question.






            share|cite|improve this answer






















            • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
              – Brahadeesh
              Aug 9 at 20:10










            • @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
              – Dietrich Burde
              Aug 10 at 8:51











            • Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
              – Dietrich Burde
              Aug 10 at 9:10











            • Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
              – Brahadeesh
              Aug 10 at 10:07










            • Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
              – JenaJ
              Aug 10 at 18:33












            up vote
            1
            down vote










            up vote
            1
            down vote









            This number is the sum of three cubes, i.e.,
            $$
            113079=J^3+W^3+J'^3,
            $$
            with (large) integers $J,W,J'$.



            Reference: This MO-question.






            share|cite|improve this answer














            This number is the sum of three cubes, i.e.,
            $$
            113079=J^3+W^3+J'^3,
            $$
            with (large) integers $J,W,J'$.



            Reference: This MO-question.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 10 at 8:55

























            answered Aug 9 at 15:45









            Dietrich Burde

            74.8k64185




            74.8k64185











            • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
              – Brahadeesh
              Aug 9 at 20:10










            • @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
              – Dietrich Burde
              Aug 10 at 8:51











            • Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
              – Dietrich Burde
              Aug 10 at 9:10











            • Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
              – Brahadeesh
              Aug 10 at 10:07










            • Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
              – JenaJ
              Aug 10 at 18:33
















            • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
              – Brahadeesh
              Aug 9 at 20:10










            • @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
              – Dietrich Burde
              Aug 10 at 8:51











            • Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
              – Dietrich Burde
              Aug 10 at 9:10











            • Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
              – Brahadeesh
              Aug 10 at 10:07










            • Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
              – JenaJ
              Aug 10 at 18:33















            This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
            – Brahadeesh
            Aug 9 at 20:10




            This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
            – Brahadeesh
            Aug 9 at 20:10












            @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
            – Dietrich Burde
            Aug 10 at 8:51





            @Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
            – Dietrich Burde
            Aug 10 at 8:51













            Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
            – Dietrich Burde
            Aug 10 at 9:10





            Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
            – Dietrich Burde
            Aug 10 at 9:10













            Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
            – Brahadeesh
            Aug 10 at 10:07




            Yikes, I misunderstood your answer completely. My apologies, I’ll be more careful from now on.
            – Brahadeesh
            Aug 10 at 10:07












            Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
            – JenaJ
            Aug 10 at 18:33




            Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
            – JenaJ
            Aug 10 at 18:33


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