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Elan Coil [88]
3 years ago
5

The image of A(-1, 1) under a reflection is A' (-1, -1). Which reflection produces the

Mathematics
1 answer:
Elenna [48]3 years ago
7 0

Answer: C, Reflection in the x-axis

Step-by-step explanation:

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Simplify 30x-140-(x-4)
balu736 [363]

Answer:

x = 4.7

Step-by-step explanation:

30x - 140 -(x - 4)

Multiply the -1 into (x - 4)

30x - 140 - x + 4

Add/subtract

29x - 136 = 0

      +136

29x = 136

\frac{29x}{29} = \frac{136}{29}

x = 4.689655172413793

8 0
3 years ago
Which absolute value function defines this graph?
iogann1982 [59]
If you describe the graph as a V with its vertex point on (0,0), then you are either describing the graph of y = |x| or y = 3|x|. The only difference is that the coordinates of y=3|x| is multiplied thrice the coordinates of y=|x|. In other words, this graph is thrice wider than the y=|x|.

Furthermore, <span>y = -3|x| is a downward V, and
</span>x = |y| and <span>x = 3|y| looks like a V rotated side-ward,</span><span>
</span>
5 0
3 years ago
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
Evaluate 15/r-1 when r=5
DochEvi [55]

Answer: 15/5=3 3-1= 2

The answer would be 2. :P



6 0
3 years ago
Read 2 more answers
Wayne used the diagram to compute the distance from Ferris, to Dunlap, to Butte. How much shorter is the distance directly from
OLEGan [10]

Answer:

10 mi

Step-by-step explanation:

In this question you have to use Pythagorus Theorem. a²+b²=c²

a = 15

b = 20

So, 15²+ 20² = c²

= 225 + 400 = 625

So Ferris to Butte is \sqrt{625} which is 25.

So now you would do 15 + 20 to find out Wayne's distance. 35 - 25 = 10.

Hope it made sense and I didn't waffle

7 0
2 years ago
Read 2 more answers
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