The half-life of a radioactive substance is 15 years. Determine algebraically, to the nearest year, how long it will take for 1/10 of this substance to remain.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.

answer: 1inch=7days  

12inch=12*7=84days

step-by-step explanation: