Kody Whitaker
2022-09-22
Answered

How do you write the equation $y-5=6(x+1)$ in slope intercept form?

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devilvunga

Answered 2022-09-23
Author has **14** answers

The slope-intercept form of a linear equation is: y=mx+b

Where m is the slope and b is the y-intercept value.

First, expand the terms in parenthesis on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$y-5={6}(x+1)$

$y-5=({6}\times x)+({6}\times 1)$

$y-5=6x+6$

Now, add 5 to each side of the equation to solve for y while keeping the equation balanced:

$y-5+{5}=6x+6+{5}$

$y-0=6x+11$

$y={6}x+{11}$

Where m is the slope and b is the y-intercept value.

First, expand the terms in parenthesis on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$y-5={6}(x+1)$

$y-5=({6}\times x)+({6}\times 1)$

$y-5=6x+6$

Now, add 5 to each side of the equation to solve for y while keeping the equation balanced:

$y-5+{5}=6x+6+{5}$

$y-0=6x+11$

$y={6}x+{11}$

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I want to know how many ways there are to choose $l$ elements in order from a set with $d$ elements, allowing repetition, such that no element appears more than 3 times. I've thought of the following recursive function to describe this:

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It isn't too hard to evaluate this function by hand for very small l or by computer for small l, but I would like to find an explicit form. However, while I know how to turn recurrence relations with only one variable into explicit form by expressing them as a system of linear equations (on homogeneous coordinates if a constant term is involved) in matrix form, I don't know how a four variable equation such as this can be represented explicitly. There's probably a simple combinatorical formulation I'm overlooking. How can this function be expressed explicitly?

$C({n}_{1},{n}_{2},{n}_{3},0)=1$

$C({n}_{1},{n}_{2},{n}_{3},l)={n}_{1}C({n}_{1}-1,{n}_{2},{n}_{3},l-1)+{n}_{2}C({n}_{1}+1,{n}_{2}-1,{n}_{3},l-1)+{n}_{3}C({n}_{1},{n}_{2}+1,{n}_{3}-1,l-1)$

The number of ways to choose the elements is then $C(0,0,d,l)$. Clearly there can be at most ${3}^{l}$ instances of the base case $C({n}_{1},{n}_{2},{n}_{3},0)=1$. Additionally, if ${n}_{i}=0$, that term will not appear in the expansion since zero times anything is zero.

It isn't too hard to evaluate this function by hand for very small l or by computer for small l, but I would like to find an explicit form. However, while I know how to turn recurrence relations with only one variable into explicit form by expressing them as a system of linear equations (on homogeneous coordinates if a constant term is involved) in matrix form, I don't know how a four variable equation such as this can be represented explicitly. There's probably a simple combinatorical formulation I'm overlooking. How can this function be expressed explicitly?

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