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n200080 [17]
3 years ago
10

How many possible combinations are there for the values of l and ml when n = 5?.

Mathematics
2 answers:
jasenka [17]3 years ago
7 0

5 (l) and 25 (ml)

HOPED I HELPEDDD

adoni [48]3 years ago
6 0

Answer:

or the last part of your question..."shell" refers to the principle quantum number, n, and "subshell" refers to the angular momentum, or the subshell/orbitals or s/p/d/f. So, the fourth shell would be when n=4, and the orbital/subshell can be l=0,1,2,3 for the "4th shell"

Step-by-step explanation:

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Hich of the following is a true statement about perpendicular lines?
IRISSAK [1]

Answer:

C.

perpendicular lines intersect in at least two points..

ok im not sure sorry

3 0
3 years ago
Read 2 more answers
(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
3 years ago
Write an expression for the product of 72 and a number
xz_007 [3.2K]
72x

That's is the unknown number multiplied by 72.
8 0
3 years ago
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What is the area of the red region?
Nitella [24]

Answer:

≈ 57.13cm^{2}

Step-by-step explanation:

Area of Circle:

A = π r^{2}

π (4)^{2}

π (16)

≈ 50.27cm^{2}

Area of Square:

A = l(w)

8(8) = 64 cm^{2}

Area of each corner:

64 - 50.27 = 13.73cm^{2}

13.73 ÷ 4 ≈ 3.43cm^{2}

Area of Red:

50.27 + (3.43(2)) ≈ 57.13cm^{2}

3 0
3 years ago
What is the image of the point (3,4) after a rotation of 90 degree counterclockwise about the origin
jeka57 [31]

Answer:

(- 4, 3 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x ) , then

(3, 4 ) → (- 4, 3 )

6 0
3 years ago
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