The half life of Abigailium-55 is 98 days. If there was 115g of the substance initially, algebraically determine the number of days until there is less than 15g remaining.

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step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!