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alexandr1967 [171]
3 years ago
5

Finding equation that passes through (-3, -7) and (0,8)

Mathematics
1 answer:
wel3 years ago
8 0
Find slope first
(8+7)/(0+3) = 15/3 = 5
Y = 5x + b
Plug in the point
-7 = 5(-3) + b, b = 8
Solution: y = 5x + 8
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Help!!!!!! Find x and y so the quadrilateral is a parallelogram.
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Answer:

Conclusion:

  • The value of x = 7
  • The value of y = 4

Step-by-step explanation:

Given

  • The parallelogram PQRS

We know that the diagonals of a parallelogram bisect each other.

Therefore,

  • RT = TP

Given RT = x and TP = 5x-28, so

x = 5x-28

5x = x+28

5x-x = 28

4x = 28

divide boh sides by 4

4x/4 = 28/4

x = 7

Thus, the value of x = 7

Similarly,

QT = TS

Given QT = 5y and TS = 2y+12, so

5y = 2y+12

5y-2y = 12

3y = 12

divide both sides by 3

3y/3 = 12/3

y = 4

Thus, the value of y = 4

Conclusion:

  • The value of x = 7
  • The value of y = 4
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What is the measure of <4 if the measure of <1 = 110°?
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1. Derive the half-angle formulas from the double
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1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following <em>trigonometric</em> formulae are used to derive the <em>half-angle</em> formulas:

sin² θ / 2 + cos² θ / 2 = 1                      (1)

cos θ = cos² (θ / 2) - sin² (θ / 2)           (2)

First, we derive the formula for the sine of a <em>half</em> angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a <em>half</em> angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a <em>half</em> angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in <em>rectangular</em> form to <em>polar</em> form are obtained by <em>trigonometric</em> functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in <em>polar</em> form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in <em>rectangular</em> form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the <em>linear</em> function y = 5 · x - 8, we proceed to use the following <em>substitution</em> formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

#SPJ1

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Answer:

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Step-by-step explanation:

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