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3 years ago

Round to the nearest tenth for 69.08, 43.96, 39.564 please show how to do it

2 answers:

5 0

Answer: 69.1, 44.0,39.6

explanation: you look at the value next to the tenths place which is the hundredths and if the number is above a five you add one to the tenth place but if the number is under 5 you keep the tenths place the same

Wittaler [7]3 years ago

5 0

Answer:

69.08 ⇒ nearest tenth

To round 69.08 to the nearest tenth, 0 is in the tenths place and 8 is in the hundredths place. Since the digit in the hundredths place (8) is greater than 5, add 1 to the digit in the tenths place (0).

1 + 0 = 1, so the digit in the tenths place becomes 1 & the digit in the hundredths place is dropped.

∴ 69.08 ≈ 69.1 (corrected to the nearest tenth)

43.96 ⇒ nearest tenth

To round 43.96 to the nearest tenth, 9 is in the tenths place and 6 is in the hundredths place. Since the digit in the hundredths place (6) is more than 5, add 1 to the digit in the tenths place (9), add 1 to the digit in the tenths place (9). This will also add 1 to the whole number part and make the fractional part 0.

1 + 9 = 10

∴ 43.96 ≈ 44 (corrected to the nearest tenth)

39.564 ⇒ nearest tenth

To round 39.564 to the nearest tenth, 5 is in the tenths place and 6 is in the hundredths place. Since the digit in the hundredths place (6) is more than 5, add 1 to the digit in the tenths place (5), add 1 to the digit in the tenths place (5).

1 + 5 = 6, so the digit in the tenths place becomes 6 & the digit in the hundredths and thousandths place is dropped.

∴ 39.564 ≈ 39.6 (corrected to the nearest tenth)

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3 years ago

Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

3 0

4 years ago

Round to the nearest tenth for 69.08, 43.96, 39.564 please show how to do it ​

Round to the nearest tenth for 69.08, 43.96, 39.564 please show how to do it