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3 years ago

Adding and subtracting complex numbers question

Mathematics

1 answer:

Darina [25.2K]3 years ago

8 0

True because i² = -1 and a and b are real

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- What is the range of the following set of points: {(3, 4), (7, - 2), (11, – 8)}?

konstantin123 [22]

Answer:

B) {-8,-2,3,4,7,11}

Step-by-step explanation:

range is y

3 0

3 years ago

Looking for the answer to this

SashulF [63]

<h2>Required Answer :-</h2>

Given system of linear equations are ,

x + y = 2
-3x + y = 2

And we need to find the Solution of the linear equation . So let's Firstly number the equations .

$\large{\red{\frak{ Given}}}\begin{cases} \sf x + y = 2\qquad ..........(i) \\\\\sf -3x + y = 2 \qquad ..........(ii) \\\end{cases}$

→ Multiplying equⁿ (i) by 3 ,

=> 3 ( x + y ) = 2*3

=> 3x + 3y = 6

→ Adding the two equations ,

=> 3x + 3y -3y + y = 6 + 2

=> 4y = 8

=> y = 8/4

=> y = 2

→ Put y = 2 in (i) ,

=> x + y = 2

=> x + 2 = 2

=> x = 2- 2

=> x = 0

<h3>★ Hence the required solution is ( 0 , 2 ) .</h3>

7 0

3 years ago

Question 4(choose A, B, C, or D)

tresset_1 [31]

Answer:

I believe the answer is d

5 0

3 years ago

Please help be fast I’m on a test

Leto [7]

Hundreds because the answer is 251 so the 2 is in the hundreds

8 0

3 years ago

Finding the inverse of the function x^3+3x+1 is beyond the scope of this course. Nevertheless, you should be able to use your kn

slega [8]

Step-by-step explanation:

${x}^{3} + 3x + 1 = 1 = = = > \\ {x}^{3} + 3x = 0 = = = > \\ x({x}^{2} + 3) = 0 \\ x = 0 \: or \\ {x}^{2} + 3 = 0 = = = > \\ {x}^{2} = - 3 = = = > x 1= \sqrt{ - 3} \: or \: i \sqrt{3} \\ x2 = - \sqrt{ - 3} = = = > x2 = - i \sqrt{3}$

${x}^{3} + 3x + 1 = 5 = = = > \\ {x}^{3} + 3x - 4 = = = > \\ (x - 1)( {x}^{2} + x + 4) = 0 = = = > \\ x - 1 = 0 = = > x1 = 1 \\ {x}^{2} + x + 4 = 0 = = = > \\ x2 = - 1 + i \sqrt{3} \\ x3 = - 1 - i \sqrt{3}$

${x}^{3} + 3x + 1 = 3 = = = > \\{ x}^{3} + 3x - 2 = 0 = = = > x = 0.6$

4 0

2 years ago