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tekilochka [14]
3 years ago
7

I don’t know how to do this. Can you help please help me ?

Mathematics
1 answer:
N76 [4]3 years ago
3 0

Answer:

  • sin = -√3/2
  • cos = -1/2
  • tan = √3
  • sec = -2
  • csc = (-2/3)√3
  • cot = (√3)/3

Step-by-step explanation:

See the attached picture for a drawing of the angle and its terminal point coordinates. Those are (cos(4π/3), sin(4π/3)), so we have the following trig function values:

  sin(4π/3) = -√3/2

  cos(4π/3) = -1/2

  tan(4π/3) = sin/cos = √3

  sec(4π/3) = 1/cos = -2

  csc(4π/3) = 1/sin = -(2√3)/3

  cot(4π/3) = 1/tan = (√3)/3

_____

<em>Additional comment</em>

It helps to know that 1/√a = (√a)/a. This lets you write the ratios with a rational denominator in each case.

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Could someone help me for this?
lina2011 [118]

Answer:

x = - \frac{5}{3} , x = \frac{5}{2}

Step-by-step explanation:

to find the points of intersection equate the 2 equations , that is

7x - 15 = 10 + 12x - 6x² ( subtract 10 + 12x - 6x² from both sides )

6x² - 5x - 25 = 0 ← factor the quadratic on left side

consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 6 × - 25 = - 150 and sum = - 5

the factors are - 15 and + 10

use these factors to split the x- term

6x² - 15x + 10x - 25 = 0 ( factor the first/second and third/fourth terms )

3x(2x - 5) + 5(2x - 5) = 0 ← factor out (2x - 5) from each term

(2x - 5)(3x + 5) = 0

equate each factor to zero and solve for x

3x + 5 = 0 ⇒ 3x = - 5 ⇒ x = - \frac{5}{3}

2x - 5 = 0 ⇒ 2x = 5 ⇒ x = \frac{5}{2}

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2 years ago
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Answer:

D

Step-by-step explanation:

Hope this helped!

7 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

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Solve. Show all of your steps. 3.) 3(5 + 2) - 2 cubed​
Aleonysh [2.5K]

Answer:

4 3=64

Step-by-step explanation:

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