Wiki User
∙ 15y ago22.8 seconds
Wiki User
∙ 15y agoThe remainder is 2-p or 0.5p of the original amount.
The answer depends on 3240 WHAT: seconds, days, years?
1/4
Multiplying a function by -1 will make it a reflection of the original function across the x axis.
Put the values that you find (as the solution) back into one (or more) of the original equations and evaluate them. If they remain true then the solution checks out. If one equation does not contain all the variables involved in the system, you may have to repeat with another of the original equations.
25
THE QUESTION IS INVALID. THE HALF-LIFE OF FLUORINE-20 is 11.07 SECONDS. There will be about 0.92 of of an isotope with a half-life of 114 seconds remaining after 14 seconds. The equation for half-life decay is AT = A0 2(-T/H) where T is time and H is half-life. 2(-14/114) is equal to about 0.92. In the case of fluorine-20, 2(-14/11.07) is about 0.42.
After 1 year, 50% of the original amount of cobalt-60 will remain. This means that 50% will decay and 50% will be left. After 4 years, 6.25% of the original amount (50% of 50%) of cobalt-60 will remain.
After 4 days, only 1/16 of the original amount of gold (198/16 = 12.375) will remain.
5.4
The remainder is 2-p or 0.5p of the original amount.
1/4 of the amount that was originally there will remain. Note that the rest doesn't simply evaporate, it will decay into something else.
After 14 years, 1/16th of the original amount of cobalt-60 will remain, because 14 years is equivalent to 2.64 half-lives of cobalt-60 (14 years / 5.3 years/half-life). Each half-life reduces the amount of cobalt-60 by half, so after 2.64 half-lives, the original amount will be reduced to 1/2^2.64 which is approximately 1/16th.
There is no possible way to 'get skinny' in five seconds and remain healthy.
original remain is organisms encased in amber and i dont know if you can find it in tar
how important is for a film maker to remain true to the original story
how important is for a film maker to remain true to the original story