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2 years ago

Please Help :(

Mathematics

2 answers:

Arturiano [62]2 years ago

7 0

Answer:3 71/75

Step-by-step explanation://Give thanks(and or Brainliest) if helpful (≧▽≦)// Plz

kogti [31]2 years ago

7 0

Answer:

the answer is 3 71/75

Step-by-step explanation:

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I need help finding the angles. The answer Isn't even

mart [117]

The 3 inside angles of a triangle equal 180.

Add the 3 angles together to equal 180:

X + (5x-20) + 4x = 180

Simplify the left side by adding like terms:

10x-20 = 180

Add 20 to each side:

10x = 200

Divide both sides by 10:

x = 20

Angle A = x = 20

Angle B = 4X = 4(20) = 80

Angle C = 5x-20 = 5(20) -20 = 100-20 = 80

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2 years ago

Find the volume <br> Find the volume

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2 years ago

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7/15 divided by 3/10

Yuliya22 [10]

Answer:

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Step-by-step explanation:

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2 years ago

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).

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3 years ago

Roups

Burka [1]

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<em>NOT VALID</em>

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2 years ago