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Masteriza [31]
3 years ago
10

Find the measure of the arc or angle indicated. or find the value of x

Mathematics
1 answer:
bonufazy [111]3 years ago
5 0
#8 you do 360= 112+136+x... then you solve for x. X=112. Now you divide this by two since it is an inscribed angle. For #9 I have this very simplified work in the picture. Sorry I did not do #10

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When 3x² - 2x + 1 is subtracted from 2x² +7x + 5, the result will be?​
Karolina [17]

Answer:

-x²+9x+4

Step-by-step explanation:

Hope this helps!

4 0
3 years ago
Read 2 more answers
What is 1.3742 rounded to the nearest tenth?
REY [17]

Answer:

1.4 that is the true answer

Step-by-step explanation:

go to the place right after the decimal and then look at the number next to it.if it is greater the 5 it goes up to 1.4 and it is

6 0
2 years ago
(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
2 years ago
Could someone help me find the domain and range? Thank you
Sonja [21]
(-4,0) (-3,3) (-1,0) (0,-3) (1,-5) (3,0)
So, domain numbers are -4, -3, -1, 0, 1, 3
Range are 0, 3, -3,-5,
Just take the ordered pairs and the Xs is the domain and the Ys are the range
8 0
3 years ago
Y=10/x+5 linear or nonlinear
Volgvan

Answer: Linear

Step-by-step explanation:

The equation y=mx+b which is used for this equation is a linear function.

8 0
2 years ago
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