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andre [41]
3 years ago
13

How to solve logarithms

Mathematics
1 answer:
statuscvo [17]3 years ago
5 0
To solve<span> a logarithmic equation, rewrite the equation in exponential form and </span>solve<span>for the variable</span>
You might be interested in
What is x in 2(x+20)=90
katovenus [111]

Answer:

25

Step-by-step explanation:

2(25+20)=90. 25 plus 20 is 45 and 45 times 2 is 90

6 0
3 years ago
Read 2 more answers
How to solve this. I am very confused.
sergiy2304 [10]

To solve, we will need to plug in -5 for x in both instances.

|-5 + 2| / -5 + 2

|-3| / -3

Now, these absolute value bars may be a bit puzzling. What we have to do is take the absolute value of the number inside of the brackets. Meaning, that if the number is negative, we make it positive, and if the number is positive, then it stays positive.

3 / -3

-1

Hope this helps!! :)

6 0
3 years ago
Read 2 more answers
I just need no. a please help me to prove this. ​
OleMash [197]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    and         cos A = cos B · cos C

scratchwork:

  A + B + C = π

               A = π - (B + C)

         cos A = cos [π - (B + C)]                              Apply cos

                    = - cos (B + C)                                    Simplify

                    = -(cos B · cos C - sin B · sin C)          Sum Identity

                    = sin B · sin C - cos B · cos C               Simplify

cos B · cos C = sin B · sin C - cos B · cos C               Substitution

2cos B · cos C = sin B · sin C                                        Addition

                     2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}                                     Division

                     2 = tan B · tan C

\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}

<u>Proof LHS → RHS</u>

Given:                              A + B + C = π

Subtraction:                     A = π - (B + C)

Apply tan:                  tan A = tan(π - (B + C))

Simplify:                               = - tan (B + C)

\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)

Substitution:                        = -(tan B + tan C)/(1 - 2)

Simplify:                               = -(tan B + tan C)/-1

                                            = tan B + tan C

LHS = RHS:   tan B + tan C = tan B + tan C  \checkmark

5 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
3 years ago
Simplest form equivalent to 3y + y + 2y - 4y
bixtya [17]

Answer:

4y+6y

Step-by-step explanation

PUT ME THE BRAINLIEST PLEASE

7 0
3 years ago
Read 2 more answers
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