+0

# How much of the beginning mass remains after 70 years?

0
492
2
+564

----------------------------------

I'm not really sure how to even begin this problem? Help would be deeply appreciated! Thanks in advanced

Jun 17, 2015

#1
+28357
+15

Simply replace t by 70:

$${\mathtt{y}} = {\mathtt{40}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{70}}}{{\mathtt{40}}}}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{11.892\: \!071\: \!150\: \!027\: \!210\: \!7}}$$

y = 11.89 grams to the nearest one-hundredth of a gram

.

Jun 17, 2015

#1
+28357
+15

Simply replace t by 70:

$${\mathtt{y}} = {\mathtt{40}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{70}}}{{\mathtt{40}}}}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{11.892\: \!071\: \!150\: \!027\: \!210\: \!7}}$$

y = 11.89 grams to the nearest one-hundredth of a gram

.

Alan Jun 17, 2015
#2
+564
+10

Wow.. it was that easy?! Thanks so much Alan!

Jun 17, 2015