Find the infinite geometric sum of 80 + 10 +1.25 + to 5 decimal places past the decimal point.

first term.

a= 80

second term. a2=10

so Common ratio r=a2/a =10/80=1/8=0.125

checking for the validity of common ratio.

a2×r=10×1/8=10×0.125=1.25 =a3= Third term.

Hence it is true.

Now 4th Term =1.25/8=0.15625

It can be observed that 4TH term Has digit that is 5 places past the decimal point.

Hence, 0.15625 is the terminating number as per the question.

So the required sum is.

S=a+a2+a3+a4

->S=80+10+1.25+0.15625

->S= 91.40625

Hope it helps...

Regards,

Leukonov/Olegion.

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950