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Svetradugi [14.3K]
3 years ago
14

A classic counting problem is to determine the number of different ways that the letters of “dissipate” can be arranged. Find th

at number
Mathematics
1 answer:
alexdok [17]3 years ago
6 0

Answer:

90720 ways

Step-by-step explanation:

Since there are 9 letters, there are 9! ways to arrange them. However since there are repeating letters, we have to divide to remove the duplicates accordingly. There are 2 ‘s’ and 2 ‘i’ hence:

Number of way to arrange ‘dissipate’ = 9! / (2! x 2!) = 90720 ways

Hence there are 90720 ways to have the number of dissipate in the letter.

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I'll send a pic of the questions
Gnoma [55]

The initial expression is:

-10+\frac{r}{5}=1

So we can solve for r so:

\begin{gathered} \frac{r}{5}=1+10 \\ r=11\cdot5 \\ r=55 \end{gathered}

7 0
1 year ago
there were twenty one adults in line at a movie theater.That is three times the number of children in line. how many children we
V125BC [204]
You would do 21 divided by 3 and get 7.

The answer is 7 kids are in the line.




Please vote my answer the brainiest if it was helpful!!








7 0
3 years ago
Need answers for 70-75
adoni [48]

Answer:

70. 0

71. -54

72. 12

73. 86

74. 59

Step-by-step explanation:

To evaluate an expression, substitute specific values for the variables and simplify using Order of Operations.

70. c-3d becomes

(1)-3(\frac{1}{3}) = 1-(\frac{3}{3}) = 1-1 = 0

71. x+x^3-4x^4 becomes

2+2^3-4(2)^4 = 2+8-4(16) = 2+8-64=10-64=-54

72. 5a+3a^2 becomes

5(-3)+3(-3)^2 = -15 +3(9) = -15 + 27 = 12

73. F=\frac{9}{5}C+32 becomes

F=\frac{9}{5}(30)+32=\frac{270}{5}+32 = 54+32 = 86

74.  F=\frac{9}{5}C+32 becomes

F=\frac{9}{5}(15)+32=\frac{135}{5}+32 = 27+32 = 59

8 0
3 years ago
You are to manufacture a rectangular box with 3 dimensions x, y and z, and volume v=8000. Find the dimensions which minimize the
Inga [223]

Answer:

20 by 20 by 20

Step-by-step explanation:

Let the total surface of the rectangular box be expressed as S = 2xy + 2yz + 2xz

x is the length of the box

y is the width and

z is the height of the box.

S = 2xy + 2yz + 2xz ... 1

Given the volume V = xyz = 8000 ... 2

From equation 2;

z = 8000/xy

Substituting into equation 1;

S = 2xy + 2y(8000/xy)+ 2x(8000/xy)

S = 2xy+16000/x+16000/y

Differentiating the resulting equation with respect to x and y will give;

dS/dx = 2y + (-16000x⁻²)

dS/dx = 2y - 16000/x²

Similarly,

dS/dy = 2x  + (-160000y⁻²)

dS/dy = 2x - 16000/y²

Note that at the turning point, ds/dx = 0 and ds/dy = 0, hence;

2y - 16000/x² = 0 and 2x - 16000/y² = 0

2y = 16000/x² and 2x = 16000/y²

2y = 16000/(8000/y²)²

2y = 16000×y⁴/64,000,000

2y = y⁴/4000

y³ = 8000

y =³√8000

y = 20

Given 2x = 16000/y²

2x = 16000/20²

2x = 16000/400

2x = 40

x = 20

Since Volume of the box is V = xyz

8000 = 20(20)z

8000 = 400z

z = 8000/400

z = 20

Hence, the dimensions which minimize the surface area of this box is 20 by 20 by 20.

3 0
3 years ago
Consider the function f(x) = x² + 10x + 25 for x ≥ -5.<br> What is the value of f-¹(x) when x = 4?
prisoha [69]

Answer:

-3

Step-by-step explanation:

5 0
1 year ago
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