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Oksi-84 [34.3K]
1 year ago
6

How many 5-letter words can be made using exactly 5 of the letters from

Mathematics
1 answer:
kow [346]1 year ago
3 0

The total number of ways by words are formed is 35 ways.

According to the statement

We have given that the 5 letters and we have to make word from them and the letters are repeated as equal to letters in given words.

And we have to find the possible ways.

So, The given words are:

TEXAS and MEXICO

Here X = 2 and E = 2 and all other words are one time used words.

We can find possible ways by use of combination and permutation.

So,

Total number of ways = 3C_{1} ^{5} + 2C_{2} ^{5}

here 3 because three letters are not repeatable and 2 letters are repeated for 2 times.

So,

Total number of ways = 3C_{1} ^{5} + 2C_{2} ^{5}

Total number of ways = 3(\frac{5!}{1!} ) + 2(\frac{5!}{2!*3!} )

Total number of ways = 3(5) + 2(10)

Total number of ways = 15 + 20

Total number of ways = 35.

So, The total number of ways by letters are formed is 35 ways.

Learn more about combination and permutation here

brainly.com/question/11732255

#SPJ1

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write a function g whose graph represents a translation 2 units to the right followed by a horizontal stretch by a factor or 2 o
Alenkinab [10]
\bf f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}\\\\
f(x)=&{{  A}} \left|{{ B }}x+{{  C}}  \right|+{{  D}}
\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see,
 
\bf f(x)=|x| \implies \begin{array}{lllccll}
f(x)=&1|&1x&+0|&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}

so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.

the smaller A is, the wider it opens, the larger it is, the more it shrinks.
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