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3 years ago

e="Expand $(a + 2)(3a^2 + 12)(a - 2).$" alt="Expand $(a + 2)(3a^2 + 12)(a - 2).$" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

sashaice [31]3 years ago

5 0

Answer:

3 $a^{4}$ - 48

Step-by-step explanation:

Given

(a + 2)(3a² + 12)(a - 2)

= (a + 2)(a - 2)(3a² + 12) ← expand the first pair of parenthesis using FOIL

=(a² - 4)(3a² + 12) ← expand using FOIL

= 3 $a^{4}$ + 12a² - 12a² - 48 ← collect like terms

= 3 $a^{4}$ - 48

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(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

Match the expressions with rational exponents to their simplified forms.

Misha Larkins [42]

do it yourself

Step-by-step explanation:

6 0

2 years ago

Find the measures of the complementary angles that satisfy each case. One of the angles is 3 times larger than the other.

Nookie1986 [14]

Answer:

22.5° and 67.5°

Step-by-step explanation:

The sum of complementary angles equal 90°.

Given that one of the complementary angles is 3 times larger than the other, let "x" represent the other angle.

Thus, the following expression can be written to represent this case:

$x + 3x = 90$

Solve for x

$4x = 90$

Divide both sides by 4

$\frac{4x}{4} = \frac{90}{4}$

$x = 22.5$

The measure of the complementary angles are:

x = 22.5°

3x = 3(22.5) = 67.5°

6 0

3 years ago

Every 125 pages that Raul reads during his summer reading challenge, he will collect 5 book points to use at his local library.

monitta

Answer:

The answer is " $\frac{1}{25}$ "

Step-by-step explanation:

They might claim that perhaps the ratio to Raul would've been: $\frac{y}{x} = \frac{5}{125} = \frac{1}{25}$ per each 125(x) page, which he reviewed throughout a summer competition.

Raul's cost per unit for each paragraph reads is $\frac{1}{25}$ page points, that's why the model a link with the same unit rate. please find the graph in the attached file.