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Kazeer [188]
3 years ago
5

Which operations is the following set closed under: {2, 4, 6, ....}?

Mathematics
2 answers:
diamong [38]3 years ago
5 0
Multiplication 
Hope that helps :) <span />
GREYUIT [131]3 years ago
4 0

Answer: The answer is addition and multiplication.


Step-by-step explanation:  Closure property states that - for any numbers 'a' and 'b' belonging to a set A, if a*b also belongs to set A, then set A is said to be closed under the operation *.

The given set is X={2, 4, 6, . . .}. Let us check for the mathematical operations one by one.

(i) Addition: The addition of two even numbers is again an even number, so the given set is closed under addition. For example - 2+4=6, 4+6=10, etc.

(ii) Subtraction: The subtraction of two even numbers is either 0, an even number or a negative even number. So, X is not closed under subtraction. For example- 2-2=0, 4-2=2, 4-8=-4, etc.

(iii) Multiplication: The multiplication of two even numbers is again an even number. So, the set X is closed under multiplication. For example- 2×4=8, 6×4=24, etc.

(iv) Division: The division of two even numbers may or may not be an even number. So, X is not closed under division. For ex- 2/6=1/3, 6/8=3/4, etc.

Thus, the given set is closed under addition and multiplication.

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Pavlova-9 [17]
Perimeter is sum of all the sides.
SO, perimeter of that triangle will be
1/4 + 1/4 + 3/8.
The LCM of denominators (4,4,8) = 8.

Hence writing 1/4 = 2/8.

so,
Perimeter = 2/8+2/8+3/8 = (2+2+3)/8 = 7/8.

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3 0
3 years ago
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Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!
blagie [28]

<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

8 0
1 year ago
Determine if the following system of equations has no solutions ......................
topjm [15]

Answer:

No solutions

Step-by-step explanation:

3x+4y=-4

15x+20y=-22

——————————

-5(3x+4y)=-5(-4) multiply 1st equation by -5 so you can eliminate the variables

-15x-20y=20. Modified 1st equation

15x+20y=-22.  2nd equation

0+0=-2. Added two equations together

since 0 does not equal -2, there is no solution.

8 0
2 years ago
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shutvik [7]

Answer: 8

Step-by-step explanation:

Given

Mean of the numbers is 8

If the ratio is 1:1:2:2:3:3

Suppose the numbers are x, x, 2x, 2x, 3x, 3x

Write the mean of the numbers

\Rightarrow \dfrac{x+x+2x+2x+3x+3x}{6}=8\\\\\Rightarrow \dfrac{12x}{6}=8\\\\\Rightarrow x=4

So, the numbers are 4, 4, 8,8 ,12,12

As the total numbers are even, median is the sum of the middle values

\Rightarrow \text{Median=}\dfrac{8+8}{2}\\\\\Rightarrow \text{Median=}8

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Answer:

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Step-by-step explanation:

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