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2 years ago

What does (+6) - (-2) =

Mathematics

1 answer:

abruzzese [7]2 years ago

5 0

Answer:

Step-by-step explanation:

Hope that helps :)

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A sailboat travels a distance of 2 1/2 miles in 1/6 of an hour. Which complex fraction represents the unit rate in miles per hou

Nikitich [7]

At first glance, prolly c

3 0

2 years ago

The amount of an invoice is $1000, with terms 2/10, n45. The amount to be paid within the discount period is:

just olya [345]

Answer:The amount to be paid within the discount period is $980

Step-by-step explanation:

The term 2/10, n45

means that 2% discount is offered if the amount of invoice is paid within 10 days from the date of invoice .

n/45 means that the buyer will pay the full amount within 45 days if he /she does not pay within the discount period of 10 days .

Now if the amount is to be paid within the discount period, the amount will be

Amount to be paid= $1000- 2% of 1000

= $1000-20

=$980

6 0

3 years ago

PLEASEEE HELP AND SHOW WORK

Verdich [7]

I mean it should be 7 cause 2 angles are congruent.

6 0

2 years ago

If f(x) = 2x and g(x) = 5x + 2, what is (f + g)(x) when x = 3?

Minchanka [31]

Answer:

f÷g=2x+5x+2=7x+2

x=3

7×3+2=21+2=23

8 0

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).

4 0

3 years ago