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3 years ago

Solve the system of equations below x-2y=3 2y-3y=9

1 answer:

Crank3 years ago

6 0

First you solve 2y-3y=9.
$2y - 3y = 9 \\ subtract \: 2y \: and \: 3y \\ - y = 9 \\ divide \: by \: - 1 \\ y = - 9$
Now that you have y, then you substitute y into x-2y=3 to find X

$x - 2( - 9) = 3 \\ multiply \: - 9 \times - 2 \\ x + 18 = 3 \\ subtract \: 18 \: on \: both \: sides \\ x = - 15$
Your solution is (-15, -9)

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Let f(x)=3x+5 and g(x)=x^2 find (f-g)(x)

dlinn [17]

Answer:

(f - g)(x) = -x² + 3x + 5

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Function Notation
Combining Like Terms

Step-by-step explanation:

Step 1: Define

f(x) = 3x + 5

g(x) = x²

(f - g)(x) is f(x) - g(x)

Step 2: Find (f - g)(x)

Substitute: (f - g)(x) = 3x + 5 - x²
Rewrite: (f - g)(x) = -x² + 3x + 5

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2 years ago

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1 year 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).

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2 years ago

Are the following expressions equivalent?

DaniilM [7]

12x +13 equal to or not equal to 12x - 6 + 8 or 12x +2

No, these expressions are not equivalent.
Hope this helps!

3 0

3 years ago

Read 2 more answers

6c + 7c + 5y + 2c + 7y Thank u ahead of time:)

Luda [366]

Answer:

15c+12y

Step-by-step explanation:

6c + 7c + 5y + 2c + 7y

13c+2c +12y

15c+12y

have a great day!

6 0

2 years ago