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NARA [144]
3 years ago
5

Let A {x ∈ N : 3 ≤ x ≤ 13}, B {x ∈ N : x is even}, and C {x ∈ N : x is odd}.

Mathematics
1 answer:
murzikaleks [220]3 years ago
7 0

Answer: The answer is as follows:

Step-by-step explanation:

From the given information, we have

Let A {x ∈ N : 3 ≤ x ≤ 13}, B {x ∈ N : x is even}, and C {x ∈ N : x is odd}

A = {3,4,5,6,7,8,9,10,11,12,13}

B = {2,4,6,8,10,12,14,..........,....}

C = {1,3,5,7,9,11,13,15,17,...........,}

(a) A∩B = {4,6,8,10,12}

(b) A∪B = {3,5,7,9,11,13 and x ∈ N : x is even}

(c) B∩C = {∅}

(d) B∪C = { x ∈ N : x is even and x ∈ N : x is odd}

Example of sets A and B when A ∩ B = {3, 5} and A ∪ B = {2, 3, 5, 7, 8}

A = {2,3,5,7,8} and B = {3,5}

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Point p lies on the directed line segment from A (2,3) to B (8,0) and portions the segment in the ratio 2 to 1
Arturiano [62]

Answer:

The x-coordinate of point P is 6

Step-by-step explanation:

we have

A (2,3) and B (8,0)

we know that

Point P portions the segment AB in the ratio 2 to 1

so

AP=\frac{2}{3} AB

and

AP_x=\frac{2}{3} AB_x

where

AP_x represent the distance between the points A and P in the x-coordinates

AB_x represent the distance between the points A and B in the x-coordinates

AB_x=8-2=6\ units

AP_x=\frac{2}{3} (6)=4\ units

The x-coordinate of P is equal to

P_x=A_x+AP_x

where

A_x represent the x-coordinate of A

substitute the values

P_x=2+4=6

therefore

The x-coordinate of point P is 6

6 0
3 years ago
Which method can you use to calculate the y-coordinate of the midpoint of a vertical line segment with endpoints at(0,0) and (0,
ahrayia [7]

Answer:

(0, - 6)

Step-by-step explanation:

Since this is a vertical line then the midpoint of the y- coordinates is

\frac{1}{2}(0 - 12) = - 6

⇒ midpoint = (0, - 6 )


7 0
3 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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