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ser-zykov [4K]
3 years ago
15

PLEASE HELP!!!!

Mathematics
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

1 is smaller than ten

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A cheese box is shaped like a
prohojiy [21]

Answer:

yes because it is a triangle it goes into the box that is equal to 1/4 of its size

Step-by-step explanation:

8 0
3 years ago
The seconds hand of an analog clock completes an entire rotation in 60 seconds, and the length of the seconds hand is 5 inches.
zaharov [31]

Answer:

195

Step-by-step explanation:

3 x 60 = 180

180 + 15

195

5 0
3 years ago
Read 2 more answers
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
My brother needs help and I have load of hw plz help
Ilya [14]
The ratio indicates that there are 8 dolls for every 7 teddy bears. That means that if there is 16 dolls, then there is 14 teddy bears. And that if there is 24 dolls, then there is 21 teddy bears. You could continue this pattern, or you could divide and multiply. If you have 160 dolls, then do 160/8 to simplify the ratio. You get 20. So if there is (8x20) dolls, then there is (7x20) teddy bears. 7x20 is 140, so there are 140 teddy bears.
5 0
3 years ago
Read 2 more answers
4x + (-3) = 5x + 3 what does x equal
lianna [129]

Answer:

x = -6

Step-by-step explanation:

4x -3 = 5x +3

-4x        -4x

-3 = x + 3

-3        -3

-6 = x

8 0
3 years ago
Read 2 more answers
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