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tamaranim1 [39]
3 years ago
14

What is 40% in fractions

Mathematics
2 answers:
IrinaVladis [17]3 years ago
6 0

4/10, which can be reduced to 2/5

Snowcat [4.5K]3 years ago
3 0

Answer:

2/5

Step-by-step explanation:

40 % = 40/100=4/10=2/5

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62.5/100 and 0.625 are both equivalent.
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Use counting to determine the whole number that corresponds to the cardinality of these sets:______. (a) A= (xl x € Nand 20.<
Harlamova29_29 [7]

Answer:

Step-by-step explanation:

Cardinality of a set is defined as number of element in a set. It is represented as n(X) where X is any set.

a) Given the set A =(x l x € N and 20.< x<27). According to the set, the set contains the values of natural numbers between 20 and 27. The values are 21, 22, 23, 24, 25 and 26

A = (21, 22, 23, 24, 25, 26)

According to the set A, it can be seen that there are 6 elements in the set, <em>this means n(A) = 6.</em>

b) Given B=(x | xeN and x+1=x)

Since natural numbers starts from 1, the first element in the set is 2 i.e 1+1

The elements of the set B = (2, 3, 4, 4...)

<em>The number of whole number in the set is therefore infinite. </em>

c) For the set C={x l xe N and (x- 1)(x - 9) = 0)

We need to get the root of the equation  (x- 1)(x - 9) = 0

x-1 = 0 and x-9 = 0

x = 1 and x = 9

Hence the element C = (1,9)

<em>The number of whole number in the set is n(C) = 2</em>

d)  D={xlx E N, H X 5 100, and x is divisible by both 5 and 8)

Given the value of x between 5 and 100, the values of x divisible by 5 = (10, 15, 20, 25, 30, 35, 40, 45, 50 , 55, 60 , 65, 70, 75, 80 85, 90, 95)

values of x divisible by 8 = (16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96)

<em>The total number of elements in both set = 29 i.e n(D) = 29 </em>

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Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(4​,5​), ​(-6​,-6​), and ​(-14​,2​
Lilit [14]

Answer:

  (-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

  (x -4)^2 +(y -5)^2 = 10^2

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  x^2 +y^2 -8x -10y = 59

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The epicenter is 5 units from Y, so lies on the circle ...

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  x^2 +y^2 +12x +12y = -47

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The epicenter is 13 units from Z, so lies on the circle ...

  (x +14)^2 +(y -2)^2 = 13^2

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Subtracting the second equation from each of the other two, we get ...

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  -20x -22y = 106 . . . . eq1 -eq2

  (x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

  16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

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