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2 years ago

Jarod read 36 pages out of his 180 page book.

1 answer:

4 0

Data:
Read pages = 36
Total pages = 180
<span>The ratio of pages read = ?

Solving:
</span> $The\:ratio\:of\:pages\:read = \frac{read\:pages}{total\:pages}$
$The\:ratio\:of\:pages\:read = \frac{36}{180} \frac{\div36}{\div36} \to \boxed{\boxed{The\:ratio\:of\:pages\:read = \frac{1}{5}}}$ <span>
</span>

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What is the effect on the graph of f(x)=1xf(x) = \frac{1}{x}f(x)=x

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Answer:

Step-by-step explanation:

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2 years ago

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The answer would be 0.5 gallons.

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2 years ago

A neighborhood was given a vacant lot in the shape of a rectangle on which to build a park. The neighborhood is

AnnZ [28]

Answer:

The answer to your question is below

Step-by-step explanation:

Formula

Triangle's Area = 1/2 bh/2

Trapezoid's area = 1/2 (b1 + b2)h

Parallelogram's area = bh

Rectangle's area = bh

Considering the formulas, we can conclude that:

a) The first choice is true, both formulas have 1/2 in.

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2 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).

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2 years ago

It took 12 men 5 hours to build an airstrip. Working at the same rate, how many additional men could have been hired in order fo

motikmotik

Answer:

Step-by-step explanation:

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4 months ago