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Mkey [24]
2 years ago
10

Divide (- 6 + i)/(1 - 2i) =

Mathematics
2 answers:
hichkok12 [17]2 years ago
5 0

\dfrac{-6+i}{1-2i}=\dfrac{(-6+i)(1+2i)}{(1-2i)(1+2i)}=\dfrac{-6-12i+i-2}{1+4}=\dfrac{-8-11i}{5}=-\dfrac{8}{5}-\dfrac{11}{5}i

Kaylis [27]2 years ago
4 0

~~~\dfrac{(-6 +i)}{(1- 2i)}\\\\\\=\dfrac{(-6+i)(1+2i)}{(1-2i)(1+2i)}\\\\\\=\dfrac{-6-12i+i+2i^2}{1-(2i)^2}\\\\\\=\dfrac{-6-11i-2}{1+4}~~~~~~~~~~~~~;[i^2=-1]\\\\\\=\dfrac{-8-11i}{5}\\\\\\=-\dfrac 85-\dfrac{11}{5}i

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bearhunter [10]

Answer:

A) sin θ = 3/5

B) tan θ = 3/4

C) csc θ = 5/3

D) sec θ = 5/4

E) cot θ = 4/3

Step-by-step explanation:

We are told that cos θ = 4/5

That θ is the acute angle of a right angle triangle.

To find the remaining trigonometric functions for angle θ, we need to find the 3rd side of the triangle.

Now, the identity cos θ means adjacent/hypotenuse.

Thus, adjacent side = 4

Hypotenuse = 5

Using pythagoras theorem, we can find the third side which is called opposite;

Opposite = √(5² - 4²)

Opposite = √(25 - 16)

Opposite = √9

Opposite = 3

A) sin θ

Trigonometric ratio for sin θ is opposite/hypotenuse. Thus;

sin θ = 3/5

B) tan θ

Trigonometric ratio for tan θ is opposite/adjacent. Thus;

tan θ = 3/4

C) csc θ

Trigonometric ratio for csc θ is 1/sin θ. Thus;

csc θ = 1/(3/5)

csc θ = 5/3

D) sec θ

Trigonometric ratio for sec θ is 1/cos θ. Thus;

sec θ = 1/(4/5)

sec θ = 5/4

E) cot θ

Trigonometric ratio for cot θ is 1/tan θ. Thus;

cot θ = 1/(3/4)

cot θ = 4/3

5 0
3 years ago
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Ludmilka [50]

Answer:

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Step-by-step explanation:

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3 years ago
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lorasvet [3.4K]

Answer:

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Step-by-step explanation:

Let the y-intercept form of the equation of the line is,

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Here, m = Slope of the line

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Equation of the line with m = 0 will be,

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y = b

Since, line passes through (0, 6),

By substituting the values in the equation,

b = 6

Therefore, standard equation of the line passing through (0, 6) and (5, 6) will be,

y = 6

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Answer:

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Step-by-step explanation:

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