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2 years ago

Evaluate (6 +2i) / (3-i).

Mathematics

2 answers:

Anuta_ua [19.1K]2 years ago

4 0

Answer:

(6 +2i) / (3-i) = (6+2i)/(3-i) x (3+i)/(3+i) = (6+2i)(3+i) / (3-i)(3+i) = 6(3+i)+2i(3+i) / (3)^2-(i)^2 = 18+6i+6i+2i^2 / 9-i^2 = 18+12i+2i^2 / 9-i^2 = 6+12i+2i^2 / 3-i^2 = 2+12i+2i^2 / 1-i^2 = 2+12i+2 / 1 = 4+12ianswer.

qwelly [4]2 years ago

3 0

Answer:

2(4 + 3i) / 5

Step-by-step explanation:

(6 + 2i) (3 + i) / (3 - i) (3 + i)

=> 18 + 12i + 2i² / 9 - i²

=> 16 + 12i / 10

=> 8 + 6i / 5

=> 2(4 + 3i) / 5

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kirill [66]

Answer:

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Step-by-step explanation:

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2 years ago

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mrs. johnson gave a sum of money to her son and daughter in the ratio 5:6. Her daughter received $2400. how muh did Mrs. Johnson

Tresset [83]

To find the total amount Mrs. Johnson gave away, you need to know how much she gave the son. Because the ratio of the money given is 5:6, the amount of money he received has to be 5/6 of the money the daughter received.

So, the son received 5/6 of 2400.
To find this, multiply 5/6 x 2400 to get 2000.

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3 years ago

Pls help due soon!!!

kvasek [131]

Answer:

B) 10 m

Step-by-step explanation:

Area is 20 and Height is 4

Ok so

AREA of a triangle is =1/2Base Times Height

work back

Area=20

Height=4

b=Base

20=1/2b*4

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20=2b

Flip equation

2b=20

Divide both sides by 2

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2 years ago

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Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat

patriot [66]

The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
= 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.