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Mice21 [21]
3 years ago
7

Does it matter which equation you use to plug the value of y into? Yes or No

Mathematics
2 answers:
malfutka [58]3 years ago
7 0

Answer:

Nope, as long as the y value is correct it won´t matter in the slightest.

Step-by-step explanation:

Hope this helps! Have an Amazing day!!

Andrej [43]3 years ago
3 0

Step-by-step explanation:

No we can use x in the equation.

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Crazy boy [7]
4.5. because ⁼3=9, and 9 divided by 2 is 4.5
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Round each to the place of the underlined digit. (1). 32.65 (The 6 is underlined), (2). 3.246 (The 4 is underlined), (3). 4.073
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Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and
ollegr [7]

Firstly, we'll fix the postions where the n women will be. We have n! forms to do that. So, we'll obtain a row like:

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The n+1 spaces represented by the underline positions will receive the men of the row. Then,

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)

Since there is no women sitting together, we must write that x_2,x_3,...,x_{n-1},x_n\ge1. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}

The equation (i) can be rewritten as:

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: \dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}

[I can write the proof if you want]

Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: m!

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n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

4 0
4 years ago
Given A is between Y and Z and YA= 14x, AZ= 10x, and YZ= 12x + 48, find AZ
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3 0
3 years ago
F(x) = x^2+x-2/x^2-3x-4
Andrej [43]

i. Domain and Range

The given function is

f(x)=\frac{x^2+x-2}{x^2-3x-4}


The domain of this function is,

x^2-3x-4\ne 0

(x-4)(x+1)\ne 0

x\ne4,xne -1


The range refers to the y-values for which x is defined. x  is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put f(x)=0

This implies that;

\frac{x^2+x-2}{x^2-3x-4}=0


This will give us

x^2+x-2=0

\Rightarrow x^2+x-2=0


\Rightarrow x^2+2x--x-2=0

\Rightarrow x(x+2)-1(x+2)=0

\Rightarrow (x+2)(x-1)=0


\Rightarrow (x+2)=0,(x-1)=0

\Rightarrow x=-2,x=1

The x-intercepts are (-2,0),(1,0)


For y-intercept, we put

x=0 to obtain;

f(0)=\frac{0^2+0-2}{0^2-3(0)-4}

f(0)=\frac{1}{2}

The y-intercept is

(0,\frac{1}{2})

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.


We divide the leading coefficient of the numerator by the leading  coefficient of the denominator.


The horizontal asymptote is y=\frac{1}{1}=1

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

x^2-3x-4=0


This implies that;

x^2+x-4x-4=0

Split the middle term

x(x+1)-4(x+1)=0

Factor

(x+1)(x-4)=0

Factor further

(x+1)=0,(x-4)=0

x=-1,x=4


The vertical asymptotes are x=-1,x=4




8 0
3 years ago
Read 2 more answers
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