A rational number stops and can be turned into a fraction and doesn’t repeat while an irrational number does the opposite
If you're multiplying by 10, you're moving the decimal point to the right, the same with any multiplication in the 10's, so 100, 1,000, etc. This is because you are making the number 10 times larger, so if you had 3.4, the 4 is in the 1/10 column so multiplying by 10 would put it in the units column, making the number 34.
On the other hand, dividing by the 10's makes the decimal point move to the left because you are making the number 10 times smaller, so if the units column is worth 1, then 10 times less than that is 1/10, which is the first decimal place.
I hope this helps! I tried to explain it well but let me know if you still don't understand or if I've confused you in any way :)
Answer:
520
Step-by-step explanation:
must be the answer..
Let

where we assume |r| < 1. Multiplying on both sides by r gives

and subtracting this from
gives

As n → ∞, the exponential term will converge to 0, and the partial sums
will converge to

Now, we're given


We must have |r| < 1 since both sums converge, so


Solving for r by substitution, we have


Recalling the difference of squares identity, we have

We've already confirmed r ≠ 1, so we can simplify this to

It follows that

and so the sum we want is

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?
Answer:
4200
Step-by-step explanation:
4280=(?)+80
So the answer is 4200