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Guest
I don't get how to finish this question, I have started it.
2^(x+3) + 2^x = 288
2^x(2^3) + 2^x = 288
2^x(2^3+1) = 288
2^x(9)=288
and then what do I do to find 'x'?
This question below I didn't get at all. COuld you show me the steps towars the answer?
Cobalt-60, which is used extensively in medical radiology, has a half life of 5.3 years. This means that if you start with 100g of cobalt-60, in 5.3 years there will be 50 g left. After another 5.3 years, there will be 25g left, and so on. The amount left, AL, at any given time is given by the equation Al=A0 (1/2) ^ t/5.3, where A0 is the intial amount and t is the time in years. What fraction of the initial amount will be left after 15.9 years?
I got this only: AL=A0 (1/2) ^ (t/5.3)
= A0= 1/2)^(15.9/5.3)
What do you do next?
Thanks for the help,
-Anna
2^(x+3) + 2^x = 288
2^x(2^3) + 2^x = 288
2^x(2^3+1) = 288
2^x(9)=288
and then what do I do to find 'x'?
This question below I didn't get at all. COuld you show me the steps towars the answer?
Cobalt-60, which is used extensively in medical radiology, has a half life of 5.3 years. This means that if you start with 100g of cobalt-60, in 5.3 years there will be 50 g left. After another 5.3 years, there will be 25g left, and so on. The amount left, AL, at any given time is given by the equation Al=A0 (1/2) ^ t/5.3, where A0 is the intial amount and t is the time in years. What fraction of the initial amount will be left after 15.9 years?
I got this only: AL=A0 (1/2) ^ (t/5.3)
= A0= 1/2)^(15.9/5.3)
What do you do next?
Thanks for the help,
-Anna