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djverab [1.8K]
2 years ago
13

Factorize 16x - 6 and show correct and full working.

Mathematics
1 answer:
Vlad1618 [11]2 years ago
8 0
\left[x \right] = \left[ \frac{3}{8}\right][x]=[​8​​3​​] totally answer
You might be interested in
2/3 of 39<br><br> 3/4 of 24cm
wolverine [178]

Answer:

The answers would be

78/3

18

Step-by-step explanation:

To find them, simply treat the word "of" as if it means multiply.

2/3 * 39 = 78/3

3/4 * 24 = 18

5 0
3 years ago
Karen is buying pens and pencils for the new school year. She wants to have no more than 25 writing utensils in all. She also wa
bezimeni [28]
For this case, the first thing we must do is define variables.
 We have then:
 x: number of pens
 y: number of pencils
 We now write the system of inequations:
 x + y  \leq  25&#10;&#10;y  \geq  (x - 3) ^ 2&#10;
 The solution to the system of inequations is given by the shaded region.
 Note: see attached image.

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
Given isosceles ABC, median BD to base AC. Prove ADB=CDB. AB=, AD=, BD=, ADB=
lesya692 [45]

Answer:

1. AB=CB                     def  of isosceles triangle

2. AD=CD DEF           of median

3. BD=DB                    reflexive property

4. ADB=CDB               SSS

Step-by-step explanation:

ummmm its just the right answer trust

5 0
3 years ago
Help me please I don’t . Get so please give me the answer
Ede4ka [16]

Answer:

1953125

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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