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Tema [17]
3 years ago
14

Finit set: 1+2+3+........+25=??

Mathematics
2 answers:
yulyashka [42]3 years ago
6 0

Answer:

Finite set: A set is said to be a finite set if it is either void set or the process of counting of elements surely comes to an end is called a finite set.

In a finite set the element can be listed if it has a limited i.e. countable by natural number 1, 2, 3, ……… and the process of listing terminates at a certain natural number N.

The number of distinct elements counted in a finite set S is denoted by n(S). The number of elements of a finite set A is called the order or cardinal number of a set A and is symbolically denoted by n(A).

Thus, if the set A be that of the English alphabets, then n(A) = 26: For, it contains 26 elements in it. Again if the set A be the vowels of the English alphabets i.e. A = {a, e, i, o, u} then n(A) = 5.

Step-by-step explanation:

Veseljchak [2.6K]3 years ago
5 0

1+2+3+4+5+6..........23+24+25=325

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Harman [31]

Answer:

x=0

Step-by-step explanation:

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3 years ago
Perform the indicated operation​ and, if​ possible simplify<br> 2/3-1/7
Fiesta28 [93]
11/21 would be the answer
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3 years ago
If f(x)=x^1/2-x and g(x)=2x^3-x^1/2-x, find f(x)-g(x)
inn [45]
X^1/2 - x
-2x^3 - x^1/2 - x

-2x^3 - 2x
i think this is it
6 0
3 years ago
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Type the correct answer in each box. (____)
trapecia [35]

Answer:

1. -16; 2. +64; 3. 16

Step-by-step explanation:

The formula for the volume of a cylinder is

V = πr²x

Data:

r = (x - 8)   mm

V = 1024π mm³

Calculations:

1. Find the cubic equation  

V = πr²h

1024π = π(x - 8)² × x

Divide each side by π

1024 = x(x - 8)²  

1024 = x(x² - 16x  + 64)

1024 =    x³ - 16x² + 64x

x³ - 16x² + 64x - 1024 = 0

2. Solve the cubic equation

The general formula for a third-degree polynomial is

f(x) = ax³ + bx² + cx + d

Your polynomial is  

f(x) = x³ - 16x² + 64x - 1024

a = 1; d = -1024

According to the <em>rational roots theorem</em>, the possible roots are

factors of d/factors of a

Factors of d = ±1, ±2, ±4, ±8, ±16, ± 32, ± 64, ±128, ±256, ±512, ±1024

Factors of a = ±1

Potential roots are x = ±1, ±2, ±4, ±8, ±16, ± 32, ± 64, ±128, ±256, ±512, ±1024

That's a lot of possibilities to check by trial and error. I will just use the one that works.

Try x = 16 by synthetic division.

16|1  -16   64  -1024

  <u>|     16     0   1024 </u>

   1     0   64        0

So,

(x³ - 16x² + 64x - 1024)/(x – 16) = x² + 64

and

(x - 16)(x² + 64) = 0

x - 16 = 0        x² + 64 =    0

     x = 16       x²         = -64

                       x          =  ±8i

There is only one real root: x = 16 mm

6 0
3 years ago
22,338 rounded to the nearest hundred
kogti [31]
22300 because the tenths place is below five
8 0
3 years ago
Read 2 more answers
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