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Fiesta28 [93]
3 years ago
13

In how many ways can the letters in the word MATHEMATICS be reshuffled so that all consonants appear together?

Mathematics
1 answer:
steposvetlana [31]3 years ago
4 0

Answer:

75600

Step-by-step explanation:

We are given that  a word MATHEMATICS

Total letters =11

M repeated 2 times

T repeated 2 times

A repeated 2 times

Total vowels=4

Let MTHMTCS=P

Total number of ways in which MTHMTCS can be arranged=\frac{7!}{2!2!}

PAEAI

Total number of ways in which PAEAI can arranged=\frac{5!}{2!}

Total number of arrangements when all consonant appear together=\frac{7!}{2!2!}\times \frac{5!}{2!}

Total number of arrangements when all consonant appear together=\frac{7!5!}{2\times 2\times 2}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1\times 5\times 4\times 3\times 2\times 1}{8}=75600

By using formula ;n!=n(n-1)(n-2)...2\times 1

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Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

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Combine any like terms in the expression. If there are no like terms, rewrite the expression.
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3 years ago
What's 666 -66+777+90- 789
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Answer:

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3 years ago
At Whole foods they mix two kinds of nuts: peanuts which cost $2.50 per kilogram and walnuts which cost $4.50 per kilogram how m
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Answer: 37 kilograms of walnuts should be mixed.

Step-by-step explanation:

Let x represent the number of kilograms of peanuts needed to make the mixture.

Let y represent the number of kilograms of walnut needed to make the mixture.

The mixture to be made is 100 kg. It means that

x + y = 100

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250 - 2.5y + 4.5y = 324

- 2.5y + 4.5y = 324 - 250

2y = 74

y = 74/2 = 37

x = 100 - y = 100 - 37

x = 63

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