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marissa [1.9K]
3 years ago
10

Can somebody please help

Mathematics
1 answer:
sdas [7]3 years ago
7 0

Answer:

next row in the x column = 8

next row in the y column  = 15

Step-by-step explanation:

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Factorise y^2 - 16<br> (y squared)
cricket20 [7]

y^2-16=y^2-4^2=(y-4)(y+4)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)

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ZanzabumX [31]

Answer: The answer will be 28


Step-by-step explanation:


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4 years ago
A bicycle was originally sold for $400 was on sale for $320. what percent was the bike marked down?
Eva8 [605]
100% of the price = $400
x % of the price = $320
we need to find x

using a proportion
100 : 400 = x : 320
100 * 320 = 400 * x
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x = 32000 / 400
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That means the new price is 80% of the previous price.
100% - 80% = 20%
and based on this we can make a conclusion that the discount was 20%

I hope it is clear to you, feel free to ask anything about it :)

6 0
3 years ago
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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:

\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}

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!

Multiplying all results:

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

4 0
3 years ago
Pls help me with this
Yanka [14]

Answer:

12

Step-by-step explanation:

f(-7) = 5-(-7)=5+7=12

4 0
3 years ago
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