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larisa86 [58]
2 years ago
15

There are 7 oak trees in Aidan's yard. The other 2 trees in his yard are maple trees.

Mathematics
1 answer:
Radda [10]2 years ago
6 0
7/9 of his trees are oak trees

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If f(x)=5x^3 and g(x)=x+1, find (f•g)(x)
alukav5142 [94]

Hello from MrBillDoesMath!

Answer:

5 x^3 + 15 x^2 + 15 x + 5 , none of the provided choices

Discussion:

f(x) = 5 x^3

g(x) = x+ 1  

=>

(f•g)(x) =

f(g(x)) =

f(x+1) =

5 * (x+1)^3 =

5 x^3 + 15 x^2 + 15 x + 5

which is none of the provided answers.

Thank you,

MrB

5 0
2 years ago
Yes or no awnser pls
kifflom [539]

Answer:

No the question iant a function

5 0
2 years ago
Geometry Help please?
wariber [46]
Answers:
The coordinates of X are (-24,0)
The scale factor is 8/3.

------------------------------------------------------

To find the scale factor, you divide the y coordinates of the points Y and V
-8/(-3) = 8/3

Notice how multiplying the y coordinate of point V gets us to the y coordinate of point Y
-3*(8/3) = -24/3 = -8

Do the same for the x coordinate of point U
-9*(8/3) = -72/3 = -24
which is the x coordinate of point X
3 0
3 years ago
How can I solve this?
anygoal [31]

A = x > -3 and x < 4

B = x >withlineunder -3 and x <withlineunder 4

C = X < - 3 or X > 4


3 0
3 years ago
(a) Let R = {(a,b): a² + 3b &lt;= 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
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