Question 729494
{{{D=V/4T}}} assumes 15 drops per cc, but you do not need to know that.
 
A. {{{V=1000}}} so {{{D=1000/4T}}} which simplifies to {{{highlight(D=250/T)}}}
 
B. We just plug the value for T into the formula and find D.
For example, for {{{T=5}}} we calculate {{{D=250/5=50}}} .
For {{{T=10}}} we calculate {{{D=250/10=25}}} .
Most values for D will not be whole numbers, so I guess you would round to the nearest whole number.
For {{{T=3}}} {{{D=250/3}}}= 83.333.... = about 83.
For {{{T=4}}} {{{D=250/4=62.5}}} , which I would round up to 63.
My table would be:
{{{matrix(2,11,T,3,4,5,6,7,8,9,10,11,12,D,83,63,50,42,36,31,28,25,23,21)}}}{{{matrix(2,6,13,14,15,16,17,18,19,18,17,16,15,14)}}} {{{matrix(2,6,19,20,21,22,23,24,13,13,12,11,11,10)}}}
 
C. The graph would look like this:
{{{graph(300,300,-3,30,-10,90,250/x)}}}
You would first plot your data points (grid paper would help):
{{{drawing(300,300,-3,30,-10,90,
grid(1),
blue(circle(3,83,0.2)),blue(circle(4,63,0.2)),
blue(circle(5,50,0.2)),blue(circle(6,42,0.2)),
blue(circle(7,36,0.2)),blue(circle(8,31,0.2)),
blue(circle(9,28,0.2)),blue(circle(10,25,0.2)),
blue(circle(11,23,0.2)),blue(circle(12,21,0.2)),
blue(circle(13,19,0.2)),blue(circle(14,18,0.2)),
blue(circle(15,17,0.2)),blue(circle(16,16,0.2)),
blue(circle(17,15,0.2)),blue(circle(18,14,0.2)),
blue(circle(19,13,0.2)),blue(circle(20,13,0.2)),
blue(circle(21,12,0.2)),blue(circle(22,11,0.2)),
blue(circle(23,11,0.2)),blue(circle(24,10,0.2)),
locate(25,-5,hours),locate(1,90,drops),
locate(6.2,90,per),locate(9.5,90,minute)
)}}} Then you would draw a smooth line that would agree with the the points.
If you had a graphing calculator or some graphing software, you could make it look better, but for practical purposes the table would be enough.