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3 years ago

If a day was a year, how many does it take to make a million years?

Mathematics

2 answers:

shutvik [7]3 years ago

7 0

1,000,000 days, and it’ll take 83,333 and 10 days for a million years in days make a million years

Tasya [4]3 years ago

6 0

It would take 2,740 days to make a million years.

1,000,000 divided by 365.

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There are 3 measures of music that are played 3 times in a song. If the first measure has 8 notes, the second measure has 16 not

Mandarinka [93]

Hi!

We know that there are three measures and each has a different amount of notes. To find the total we add them all up.
8+16+4=28.

There were 28 total notes played in the song.

I hope I helped you!
Have a wonderful day!

5 0

2 years ago

Read 2 more answers

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

2 years ago

The graph g(x) is the graph of f(x) translated (5,2,3) units (down,up,left,right) , and g(x) =(f(x-3),f(x)-5,f(x)+3,f(x-2),f(x)+

marusya05 [52]

Answer:

The graph g(x) is the graph of f(x) translated 2 units right, and g(x) = f(x-2)

Step-by-step explanation:

g(x) passes through points (0, -5) and (1, -2), then the slope of g(x) is the same as the slope of f(x), which is 3.

f(x) passes through (0, 1) and g(x) passes through (2, 1). Therefore, the graph g(x) is the graph of f(x) translated 2 units right.

f(x - c) translates f(x) c units to the right, therefore g(x) = f(x-2)

In order to check this result, we make:

f(x) = 3x + 1

f(x-2) = 3(x-2) + 1

f(x-2) = 3x - 6 + 1

f(x-2) = 3x - 5 = g(x)

3 0

3 years ago

Based on the information marked in the diagram, AABC and ADEF must be

Studentka2010 [4]

Answer:

Step-by-step explanation:

answer

8 0

2 years ago

Select the correct answer.

kkurt [141]

Answer:

The answer is A I'm not 100 percent sure

5 0

3 years ago

Read 2 more answers