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3 years ago

How does a digit in the 10 thousand place compared toa digit in the 10 place

1 answer:

6 0

For example, in the number 14,509, fourteen thousand five hundred and nine, 1 is in the ten thousands place, and zero is in the tens place.

1 4, 5 0 9
ten thousand^ ten^

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Can someone pls help? Thx

sergey [27]

Your answer would be -3/4n+3/8 because 1/4•-3/1n would be -3/4n and then 1/4•3/2 would be 3/8 so your answer would then be -3/4n+3/8.

5 0

3 years ago

18. 50 - [7 + (32 - 2)]

Nataly_w [17]

Answer:

Step-by-step explanation:

32-2=30

30+7=37

50-37=13

7 0

3 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLIEST!!! What does the degree of a polynomial expression tell you about its related polynomial functi

Art [367]

Where an, an-1,a2, a1, a0 are constants. We call the term containing the highest power of x the leading term, and we call an the leading coefficient. The degree of the polynomial is the power of x in the leading term. We have already seen degree 0, 1, and 2 polynomials which were the constant, linear, and quadratic functions, respectively. Degree 3, 4, and 5

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3 years ago

Which relationship describes angles 1 and 2? Select each correct answer. A adjacent angles B complementary angles C vertical ang

Scrat [10]

The correct answer for this question is this one:
<span>- A adjacent angles
-D supplementary angles

The relationship that describes angle 1 and angle 2 is that they are adjacent angles -- at the same time they are also supplementary angles.</span>
Hope this helps answer your question and have a nice day ahead.

3 0

3 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

3 years ago