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Liono4ka [1.6K]
3 years ago
15

Sorry this a lot.

Mathematics
2 answers:
raketka [301]3 years ago
5 0
Your
<span>-4| 3 7 -20
-12 20
______________
3 -5 0
is a bit hard to read.  Try this format:
      __________
-4  /   3    7    -20
              -12   20
      -------------------
         3    -5      0

Because the remainder is 0, we know that -4 is a root and (x+4) is a factor.  Note the new coefficients in the last row:    3   -5.  They are the coeff. of the quotient, namely, 3x-5.  So (3x-5) is a factor of the original polynomial.
      

</span>
Tom [10]3 years ago
3 0

The correct answers are,

The number -4 is a root of F(x) = 3^x2 + 7x - 20.

(x + 4) is a factor of 3x^2 + 7x - 20.

(3x^2 + 7x - 20) (x + 4) = (3x - 5)

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3 years ago
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1. Derive the half-angle formulas from the double
lilavasa [31]

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

4 0
2 years ago
Which equation represents the line passing through the points (4, −1) and (1, 0)? Select one: A. x + 3y = 1 B. 3x + y = 1 C. x −
Alekssandra [29.7K]

Answer:

Step-by-step explanation:

Gradient of a line =( y2 - y1)/x2 - x1

From the question, x1=4,y1=-1,x2=1,y2=0.

Therefore, substitute for these values in the above formula

m = 0-(-1)/1-4

= 0+1/-3

= -1/3.

Therefore, y-y1/x-x1 = -1/3

y - y1 = -1/3(x - x1)

y - (-1) = -(x-x1)/3

y+1 = - (x - 4)/3

Multiply through by 3

3y+3= -x+4

3y +x=4-3

x+3y= 1

Therefore the answer is A.

7 0
3 years ago
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