Question 596026
# 1


{{{a=2(ut-s)/(t^2)}}}


{{{at^2=2(ut-s)}}}


{{{at^2=2ut-2s}}}


{{{at^2-2ut=-2s}}}


{{{-2s=at^2-2ut}}}


{{{s=(at^2-2ut)/(-2)}}}


{{{s=(-(at^2-2ut))/(2)}}}


{{{s=(-at^2+2ut)/(2)}}}


{{{s=(2ut-at^2)/(2)}}}


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# 2

You're very close to the final answer. Reduce to get


20/6 = 10/3


So the answer is 10/3


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Let d = # of dimes and q = # of quarters


"If there were 4 more dimes than quarters", then {{{d = q + 4}}}


The total value of d dimes is 0.1d dollars

The total value of q quarters is 0.25q dollars

They must add to 3.90, so 


0.1d + 0.25q = 3.90


Multiply everything by 100 to get


{{{10d + 25q = 390}}}


Now use this equation (along with the first) to solve for d and q


{{{10d + 25q = 390}}}


{{{10(q + 4) + 25q = 390}}} ... Replace 'd' with 'q+4' (since d = q+4)


{{{10q+40+25q=390}}}


{{{35q+40=390}}}


{{{35q=390-40}}}


{{{35q=350}}}


{{{q=(350)/(35)}}} 


{{{q=10}}}


So there are 10 quarters.


{{{d = q+4}}}


{{{d = 10+4}}}


{{{d = 14}}}


and there are 14 dimes