Table Height: Solving for the Perfect Fit with Cat and Tortoise Heights

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The discussion focuses on calculating the ideal table height based on the heights of a cat and a tortoise. The initial equations suggest that the table height (h) can be derived from the heights of both animals (c for cat and t for tortoise). After manipulating the equations, it is concluded that the optimal table height is 150 cm, which is proposed to be the average of the two heights. Some participants suggest simplifying the equations for clarity, indicating that parentheses are unnecessary. The overall consensus leans towards finding a balanced height that accommodates both animals effectively.
chwala
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Homework Statement
see attached
Relevant Equations
simultaneous equations
1645441839003.png

My take; Let the cat's height = ##c## and the tortoise's height =##t## and the table's height= ##h##,
therefore,
##t+ (170-c)=h##
##c+(130-t)=h##. It then follows that,
##t+ (170-c)=c+(130-t)##
##2t+40=2c##
##c=20+t##,
therefore, on substituting into ##t+ (170-c)=h##
##t+170-20-t=h##
##h=150##cm...
your thoughts...
 
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It should be the average of both measurements, I believe.
 
chwala said:
View attachment 297417

My take; Let the cat's height = ##c## and the tortoise's height =##t## and the table's height= ##h##,
therefore,
##t+ (170-c)=h##
##c+(130-t)=h##. It then
. . .

your thoughts...
Simply add the equations (There was no need for parentheses.)

You get:

##170 + 130 = 2h ## .

etc.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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