Arrange the complex numbers in order according to the quadrant in which they appear, starting with the first quadrant.

Mathematics

1 answer:

stich3 [128]3 years ago

6 0

The quadrants and the given numbers are shown in the figure below.

Answer:
The order of the complex numbers, numbered from the first quadrant are
4 + i
-2 + 2i
-1 - 3i
3 - 4i

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Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago

Solve the Inequality <br>8p<-3

Sergeeva-Olga [200]

Answer:

p<-3/8

Step-by-step explanation:

8 0

3 years ago

What is the solution to the equation 1/4x = 5?

puteri [66]

The solution to your equation is:

Solve:
1/4x = 5

Step 1: Multiply both sides by 4
4 * (1/4x) = (4) * (5)

x = 20

Thus your answer is:

Letter Choice (C), x = 20

Hope that helps!!!

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3 years ago

Limit of x^2-81/x+9<br> As x goes toward -9

Semmy [17]

Hello,

Use the factoration

a^2 - b^2 = (a - b)(a + b)

Then,

x^2 - 81 = x^2 - 9^2

x^2 - 9^2 = ( x - 9).(x + 9)

Then,

Lim (x^2- 81) /(x+9)

= Lim (x -9)(x+9)/(x+9)

Simplity x + 9

Lim (x -9)

Now replace x = -9

Lim ( -9 -9)

Lim -18 = -18
_______________

The second method without using factorization would be to calculate the limit by the hospital rule.

Lim f(x)/g(x) = lim f(x)'/g(x)'

Where,

f(x)' and g(x)' are the derivates.

Let f(x) = x^2 -81

f(x)' = 2x + 0
f(x)' = 2x

Let g(x) = x +9

g(x)' = 1 + 0
g(x)' = 1

Then the Lim stay:

Lim (x^2 -81)/(x+9) = Lim 2x /1

Now replace x = -9

Lim 2×-9 = Lim -18

= -18

7 0

3 years ago

Solve for u 2(4u-9)=14<br> plzzzz <br> will give brailiest

lions [1.4K]

ANSWER:
u=4

EXPLANATION:
distribute the 2 and simplify.

5 0

3 years ago

Read 2 more answers