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2 years ago

12

In how many ways can you arrange six trophies on a shelf?

1 answer:

melisa1 [442]2 years ago

7 0

Answer:

720 ways

Step-by-step explanation:

Assuming order doesn't matter, there are $6!=6*5*4*3*2*1=720$ ways to arrange six trophies on a shelf.

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Which expression is equivalent to 24x + 12?

Lyrx [107]

You can factor 12 from both terms to get

$24x+12 = 12(2x+1)$

7 0

2 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 <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. 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 <em>n</em> = <em>k</em> + 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

It takes 1 1/2 hours to fill 1/2 of a tank how long will it take to fill 2 tanks completely

Likurg_2 [28]

Answer:

15

Step-by-step explanation:

hope this helps

4 0

2 years ago

How much this will be ......evaluate -4 (9)(-5)

klemol [59]

-4(9)(-5) = (-36)(-5) because -4(9) = -36. (-36)(-5) = 180

3 0

2 years ago

The lsrl after an exponential transformation is log yhat=.4785 + 1.468x. What is the exponential form of the regression?

zheka24 [161]

Answer:

$10^{0.4785+1.468x}=y$

Step-by-step explanation:

So I'm assuming when you typed "log yhat=.4785 + 1.468x", you meant to write: $log(y)=0.4785+1.468x$ . And generally a logarithm can be written in the form $log_ba=c$ which can then be rewritten as $b^c=a$ , but since the log has no base, it's assumed to be 10. So in this case you have the equation:

$log_{10}y=0.4785+1.468x$ , which can then be written in exponential form as:

$10^{0.4785+1.468x}=y$

8 0

1 year ago