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3 years ago

7

Jim Smith is a word processor. He is paid $3.50 a page. A normal job has 500 pages and requires about 250 words per page. About

how much will Jim receive for his work? 875.00 1,750.00 125,000.00 437,500.00

1 answer:

worty [1.4K]3 years ago

7 0

Answer:

The correct answer will most likely be B. 1,750.00

Step-by-step explanation:

I just multiplied 3.50 by 500 to get my answer.

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Please help will give brainlyist

Shalnov [3]

Number 11
Since the auditorium can hold 600 and the student body will be divide by four parts
The answer is
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6 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

A songwriter gets paid monthly at a rate of $150 for each song he completes. Last month he wrote 8 songs and got halfway through

DaniilM [7]

Answer:

The answer would be 1275 if they get paid for half the money on the 9th song

Step-by-step explanation:

3 0

3 years ago

The degree of - 6+ 7x - 6x2 is <br> what?

andreev551 [17]

Look it up step by step

5 0

3 years ago

The length of a rectangle is 3cm more than the width. The area is 70cm^2. Find the dimensions of the rectangle

Rom4ik [11]

Let width be x
Length would be x + 3
Area = L * W
Make an equation
(x) * (x+3) = 70
x^2 + 3x = 70
x^2 + 3x - 70 = 0
Quadratic formula
(-3 +/- rt 3^2 - 4 * 1 * (-70))/(2 * 1)
x1 = (-3 + 17)/2, x2 = (-3-17)/2
x = 7, x = -10
Dimension cannot be negative

Solution: width = 7
Length = 10

7 0

3 years ago