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klasskru [66]
3 years ago
10

A string is 5.8 meters long.

Mathematics
1 answer:
choli [55]3 years ago
4 0

Answer:

the answer is 580cm in length

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Determine whether the sequences converge.
Alik [6]
a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

Notice that

\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}

So as n\to\infty you have a_n\to0. Clearly a_n must converge.

The second sequence requires a bit more work.

\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then a_n will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1

Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k

which suggests that for all n, you have a_n\ge a_{n-1}, so the sequence is increasing monotonically.

Next, based on the fact that both a_1=\sqrt2=2^{1/2} and a_2=2^{3/4}, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}
a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}

and so on. We're getting an inkling that the explicit closed form for the sequence may be a_n=2^{(2^n-1)/2^n}, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

Therefore the second sequence must also converge (to 2).
4 0
3 years ago
How to solve this equation 15(6-g)
NikAS [45]
<span>15<span>(<span>6−g</span>)</span></span><span>=<span><span>(15)</span><span>(<span>6+<span>−g</span></span>)</span></span></span><span>=<span><span><span>(15)</span><span>(6)</span></span>+<span><span>(15)</span><span>(<span>−g</span>)</span></span></span></span><span>=<span>90−<span>15g</span></span></span><span>=<span><span>−<span>15g</span></span>+<span>90</span></span></span>
7 0
3 years ago
Please help with this question thank you very much.
DochEvi [55]

Answer:

ln 3

Step-by-step explanation:

3 ln 3 - ln 9

We can rewrite 9 as 3^2

3 ln 3 - ln 3^2

We know lnx^a = aln x

3 ln 3 - 2 ln 3

ln 3

6 0
3 years ago
a weather man is found to be correct 80% of the time ( 80 out of 100 ) if he gives 375 forecasts how many of them will be correc
Galina-37 [17]
Bro don’t click the file or the link
3 0
3 years ago
A miner descended 1200 feet into a mine. While in the mine, he then rose 900 feet and then descended 350 feet. What was the mine
solniwko [45]
We can just add all these numbers together because they are just addition and subtraction.

We assume the miner starts at 0 (the surface).

0 - 1200 (down) + 900 (up) - 350 (down) = -650 (C)
6 0
3 years ago
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