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Pavlova-9 [17]
3 years ago
9

How many different 10-letter words (real or imaginary) can be formed from the letters of the word LITERATURE?

Mathematics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

The number of words that can be formed from the word "LITERATURE" is 453600

Step-by-step explanation:

Given

Word: LITERATURE

Required: Number of 10 letter word that can be formed

The number of letters in the word "LITERATURE" is 10

But some letters are repeated; These letters are T, E and R.

Each of the letters are repeated twice (2 times)

i.e.

Number of T = 2

Number of E = 2

Number of R = 2

To calculate the number of words that can be formed, the total number of possible arrangements will be divided by arrangement of each repeated character. This is done as follows;

Number of words that can be formed = \frac{10!}{2!2!2!}

Number of words = \frac{10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1}{2 * 1 * 2 * 1 * 2 * 1}

Number of words = \frac{3628800}{8}

Number of words = 453600

Hence, the number of words that can be formed from the word "LITERATURE" is 453600

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Krutika, David and Mark share some sweets in the ratio
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Answer:

120

Step-by-step explanation:

Krutika:48

David: 48

Mark: 48÷2= 24

Because 2 : 2 : 1

48+48+24=120

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What is the value of the expression 18-2×(4+3)
Vika [28.1K]
The value would be 4.

First, you have to do 4 + 3 because it's in parenteces. This equals 7
Secondly, you have to do 2 x 7, which is 14.
Last you have to do 18 - 14 which is 4.
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Mrs.Johnsons new car has an 18 gallon gas tank. If she can travel 400 miles on one tank of gasoline, what is her miles per gallo
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3 years ago
What is the answer for 1/2 x N=1/3
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1/2* N= 1/3
⇒ N= (1/3)/ (1/2)
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3 0
3 years ago
A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for
Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

\implies h=\frac{1}{w^2}

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

= 2\times \frac{1}{w^2}(w+2w)

=\frac{6w}{w^2}

=\frac{6}{w}  

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}

=2w^2+4w^2+\frac{12}{w}

=6w^2+\frac{12}{w}

Differentiating with respect to w,

C'(w) = 12w -\frac{12}{w^2}

Again differentiating with respect to w,

C''(w) = 12 + \frac{24}{w^3}

For maxima or minima,

C'(w) = 0

\implies 12w -\frac{12}{w^2}=0

\implies 12w^3 - 12=0

w^3-1=0\implies w = 1

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

7 0
3 years ago
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