Question 305549
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Let *[tex \Large d] represent the number of dimes.


Let *[tex \Large n] represent the number of nickels.


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ +\ n\ = 35]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ = 35\ -\ d]


Each dime is worth 10 cents.  Each nickel is worth 5 cents.  And $3.30 is 330 cents, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 5n\ =\ 330]


Substitute for *[tex \Large n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 5(35\ -\ d)\ =\ 330]


Just solve for *[tex \Large d]


For the second part of the question:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 5(35\ -\ d)\ =\ 250]


And solve for *[tex \Large d]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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