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.

your answer is 23/2 - 13.5

step-by-step explanation:

7/2 evaluates to

multiply x and 4.5

multiply x and 1

the x just gets copied along.

the answer is x

x

4.5*x evaluates to 4.5x

multiply 3 and 4.5x

multiply 1 and x

the x just gets copied along.

x

4.5*x*3 evaluates to 13.5x

7/2-4.5*x*3 evaluates to

  + 8 =

the answer is

7/2-4.5*x*3+8 evaluates to

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms