Help me with this very pls. I will give you the brainliest and a thanks!! :)) Can you do 1 and 4 thanks!! :)))

Math 1 pic attached

Goshia [24]

Answer:

D.

Step-by-step explanation:

(f - g) (x) ➡ f(x) - g(x)

5x - 2 - 2x - 1 = 3x - 3

4 0

3 years ago

Please due in 2 min NO LINKS

Ulleksa [173]

Answer:

$multiplication \: by \: its \: self = > power \\ multiplication = > product \\ addition = > sum \\ division = > quotient \\ substraction = > difference \\ thank \: yo$

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

Which ones are greater than the other

irakobra [83]

1. 0.91<0.93; 0.91 is farther away from the whole number 1 so it is less than 0.93.

2. 0.5=0.50; 0.5 and 0.50 are the same, since the decimal can be written with or without the zero.

3. 1.08<1.6; 1.08 is farther away from the whole number 2, so it is less than 1.6.

6 0

2 years ago

Explain how you can use the whole number expression 248÷4 to check that your answer to exercise to is reasonable

Aleksandr-060686 [28]

I got 62 because i just dived 248/4.

4 0

3 years ago