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3 years ago

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