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jekas [21]
3 years ago
10

Please help! I will mark you as brainliest!

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0

Answer:

7/9

Step-by-step explanation:

Think of 1 as being 9/9. The difference between 9/9 and 2/9 is 7/9, which is the answer

alekssr [168]3 years ago
6 0
7/9
You’re welcome
You might be interested in
Plz help me with this
forsale [732]

Answer:

\frac{31x}{30}

Step-by-step explanation:

Before adding we require the fractions to have a common denominator

The lowest common denominator of 2, 5 and 3 is 30, thus

\frac{15x}{15(2)} + \frac{6x}{6(5)} + \frac{10x}{10(3)}

= \frac{15x}{30} + \frac{6x}{30} + \frac{10x}{30}

Add the numerators leaving the denominator

= \frac{15x+6x+10x}{30} = \frac{31x}{30}

5 0
3 years ago
Read 2 more answers
X+5/2 + x−3/3 =3 1/6
Nataly [62]
The answer is x=-1/2
3 0
3 years ago
Read 2 more answers
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
Evan buys some oranges. He gives 1,2 or 3 oranges to April. He has5 oranges left . Draw pictures and write a number sentence to
stiv31 [10]

Answer:

6, 7 or 8 oranges

Step-by-step explanation:

8 0
3 years ago
Please help me<br> Show your work
GrogVix [38]

<u>ANSWER TO PART A</u>

The given triangle has vertices J(-4,1), K(-4,-2),L(-3,-1)


The mapping for rotation through 90\degree counterclockwise has the mapping


(x,y)\rightarrow (-y,x)


Therefore

J(-4,1)\rightarrow J'(-1,-4)


K(-4,-2)\rightarrow K'(2,-4)


L(-3,-1)\rightarrow L'(1,-3)


We plot all this point and connect them with straight lines.


ANSWER TO PART B


For a reflection across the y-axis we negate the x coordinates.


The mapping is



(x,y)\rightarrow (-x,y)


Therefore

J(-4,1)\rightarrow J''(4,1)


K(-4,-2)\rightarrow K''(4,-2)


L(-3,-1)\rightarrow L''(3,-1)


We plot all this point and connect them with straight lines.


See graph in attachment







7 0
3 years ago
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