N=NOe^-kt N = amount of C-14 at time t NO = initial amount of C-14 k= 0.0001 t = time (unknown)
We know that the amount found is 30% of the original. Therefore we can simply set N to 30 and NO to 100 because 30% of 100 is 30. Really you can give any amount, but setting it at 100 is the easiest, so long as N is 30% of NO.
so if we set the problem up again using all the given values and the new ones we assigned: N= 30 NO = 100
30=100e^-0.0001(t) Remember, you are solving for t, so you have to try and manipulate the formula to get t by itself on one side. We can divide both sides by 100 and you are left with: .30=e^-0.0001(t)
You can see that you have "e" which means we can use natural log to get rid of e: e^y = x and ln(x) = y
.30=e^-0.0001(t) using the natural log, .30 = x -0.0001(t) = y
Therefore: ln (x) = y ln (.30) = -0.0001(t)
plug in the value ln(.30) into a calculator and you get: -1.20397 = -0.0001(t) *NOTE: normally it would not work if you have a negative answer because you cannot have negative years, however, there is a negative on the other side of the equation which will cancel out so it works in this case.
now to solve for t, we simply divide both sides by -0.0001 Therefore t = 12039.7 But you have to round to the nearest year so t = 12040 **NOTE: the k value given (which is the rate at which C-14 decays each year) is very small. That means that you will have a large value for time because it takes so long for the C-14 to decay every year.
We know that circumference=2*pi*r for r=7 in circumference=2*pi*7------> 14*pi in
if 360° (full circle) has a length of------------ 14*pi in 135°-----------------------------> x x=135*14*pi/360----------> x=5.25*pi in--------> 16.49 in