Find the vertex of y=x^2+8x-20
1 answer:
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The series converges to 1/(1-9x) for -1/9<x<1/9
Given the series is ∑
We have to find the values of x for which the series converges.
We know,
∑ converges to (a) / (1-r) if r < 1
Otherwise the series will diverge.
Here, ∑ is a geometric series with |r| = | 9x |
And it converges for |9x| < 1
Hence, the given series gets converge for -1/9<x<1/9
And geometric series converges to a/(1-r)
Here, a = 1 and r = 9x
Therefore, a/(1-r) = 1/(1-9x)
Hence, the given series converges to 1/1-9x for -1/9<x<1/9
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Answer:
Step-by-step explanation:
We proceed to show the procedure to calculate the given fraction into a decimal form:
1) Since numerator is less than denominator, the integer component of the decimal number is zero:
2) We multiply the numerator by 10 and find the tenth digit:
Then,
3) We multiply the fraction in 2) by 10 and find the hundredth digit:
Then,
And the remainder is:
4) We multiply the remainder by 10 and divide this result by the denominator to determine the thousandth digit:
Then,
This question asks us to write a decimal correct to 2 decimal places, which has the characteristic that is infinite periodical decimal. Then, the result correct to 2 decimal places is: