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Oksi-84 [34.3K]
2 years ago
13

PLEASE HELP, WILL GIVE BRAINLIEST + 100 POINTS

Mathematics
1 answer:
attashe74 [19]2 years ago
3 0

Answer:

<em>part A.</em><em>equation of an ellipse</em>

<em>part B.</em><em>equation of a hyperbola</em>

<em>part C.</em><em>domain of each conic section</em>

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What’s the inverse of the equation
alexgriva [62]

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change the y with the x and solve for y making it the last one

Step-by-step explanation:

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3 years ago
(sqrt3-sqrt3i)^4
Ludmilka [50]

The increasing order of the complex numbers is (√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

<h3>Absolute values of the complex numbers</h3>

The absolute values of the complex numbers are determined as follows;

(sqrt3-sqrt3i)^4 = (√3 - √3i)⁴

|z| = \sqrt{(\sqrt{3} )^2 + (\sqrt{3 }\times1 )^2} } \\\\|z| = \sqrt{6}

(-1+sqrt3i)^12 = (-1 + √3i)¹²

|z| = \sqrt{(-1)^2 + (\sqrt{3)^2} } \\\\|z| = \sqrt{4} \\\\|z| = 2

(sqrt 3-i)^6 = (√3 - i)⁶

|z| = \sqrt{(\sqrt{3})^2 + (-1)^2 } \\\\|z| = \sqrt{4} \\\\|z| = 2

(sqrt2-sqrt2i)^8 = (√2 - √2i)⁸

|z| = \sqrt{(\sqrt{2} )^2 + (\sqrt{2})^2 } \\\\|z| = 2

(sqrt2-i)^6 = (√2 - i)⁶

|z| = \sqrt{(\sqrt{2})^2 + (-1)^2} } \\\\|z| = \sqrt{3}

Increasing order of the complex numbers;

(√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

Learn more about complex numbers here: brainly.com/question/10662770

#SPJ1

3 0
2 years ago
Plz need help like now plz
nika2105 [10]
T) -7x+10y=20. M) -3x+y=-2 - ,
2(8x-5y)=35(2) -1(x+y)=-6(-)

-7x+10y=20. -3x+y=-2
16x-10y=70. -x-y=6

9x=90. -4x=4
[x=10] [x= -1]

-7(10)+10y=20. -3(-1)+y=-2
-70+10y=20. 3+y=-2
10y=90. [y= -5]
[y=9]
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3 years ago
When to use 22/7 for pi?
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You don't use it usually because teachers usually say use 3.14 as an approximation
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angazii ukuthii ngingaku siza kanjanii lahh

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