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3 years ago

Just this question I've tried but don't understand it

Mathematics

1 answer:

Marta_Voda [28]3 years ago

5 0

6+4÷2=5
5*3=15
15*10= $150cm^{3}$

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In how many ways can the letters in the word spoon be arranged?

Kisachek [45]

I search and found different answers but the nearest is 24

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2 years ago

What is the volume of this solid? Recall the formula V = Bh.

Viktor [21]

Answer:

280

Step-by-step explanation:

V=Bh

V=1/2(7x10)8 *it is a triangluar pyrmid so the base is a triangle*

V=1/2x70x8

V=35x8

V=280

Answer:280

5 0

3 years ago

Read 2 more answers

Find the number whose arithmetic square root is 0.6

slavikrds [6]

Answer: 0.36

Step-by-step explanation:

6/10 x 6.10= 36/100

5 0

3 years ago

Jenin andrea marie are picking apples andre pics three times as many pounds as maria jane picks two times as many pounds is andr

Sindrei [870]

<span>Let x=apples that Andre picks
Then x/3=apples that Maria picks
And 2x=apples that Jane picks
Soooo,
x +(x/3)+2x=840 multiply each term by 3

3x+x+6x=2520 collect like terms
10x=2520
x=252</span>

5 0

3 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).

4 0

3 years ago