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3 years ago

15

I need to translate the following phase to an algebraic expression. What variable represents 10 more than a number. I think the

correct answer is let x represent the number. But I don't understand how to get the algebraic expression.

1 answer:

serg [7]3 years ago

5 0

I don't know what it is

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Have another question for you guys it is in the image

Lerok [7]

Answer:

A because its supposed to be on the y axis

5 0

3 years ago

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

Can I please get some help

Nitella [24]

Answer:

$f(3) = 3\\g(2) = -3\\g(\frac{1}{2}) = 1\\f(2) = \frac{2}{3}\\g(-2) = 0\\f(\pi) = 2\\$

Step-by-step explanation:

The functions are given f and g using coordinates.

Whenever we will ask for f(a), we look for "a" in the x coordinate of the function f and find the corresponding value. THAT IS THE ANSWER.

If we ask for g(b), we look for "b" in the x coordinate of the function g and find the corresponding value. THAT IS THE ANSWER.

So,

$f(3) = 3\\g(2) = -3\\g(\frac{1}{2}) = 1\\f(2) = \frac{2}{3}\\g(-2) = 0\\f(\pi) = 2\\$

5 0

4 years ago

How to convert whole number as fraction

Mice21 [21]

To <u>convert an integer to a fraction</u> you must choose an expression that does not alter the numerical value. You can do this with multiples of the number, for example, for 100:

100/1 = 200/2 = 300/3 = 400/4 = 100

The numerical value is not altered and the result will always be 100

Another example, for 500:

500 = 500/1 = 1000/2 = 1500/3 = 2000/4

<h2>{ Pisces04 }</h2>

3 0

1 year ago

Read 2 more answers

The sum of a number and 16 is 23

saul85 [17]

Answer:

x = 7

Step-by-step explanation:

Let x be the number

x+16 = 23

Subtract 16 from each side

x+16-16 =23-16

x = 7

5 0

3 years ago