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3 years ago

The other one is “is” or “is not”

Mathematics

1 answer:

AleksAgata [21]3 years ago

5 0

Answer: Not

Step-by-step explanation:

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Last question helo me pls

Yuri [45]

I hope this helps you

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3 years ago

SOMEONE PLEASE HELP ME WITH THIS MATH IM STRUGGLING!

NemiM [27]

Answer:

I think 58'is right answer

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3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

2 years ago

Twice the sum of a number and 5 equals 9 as a equation

ziro4ka [17]

Hello!

I believe it looks like this..

2*x+5=9

Btw, x = 3 is your answer if you're wondering..

Hope this helps! ☺♥

7 0

3 years ago

"You figure you can ride Ghost Wind, then get on. Just bring him back to the stable when you're finished so I can rub him down."

AlekseyPX

Answer:

character vs nature.

minor

I got it right on edge 2021

5 0

2 years ago

Read 2 more answers