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3 years ago

20 points and brainlyest to who answers this first in a well format!

Mathematics

1 answer:

djverab [1.8K]3 years ago

5 0

Answer:

in the forth line of the equation it should be this

-6x+10=4x+20

-6x-4x=20-10

-10/-10x=10/-10

x=-1

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Use the quadratic formula to solve 2y² + 6y – 8 = 0.

s2008m [1.1K]

{–4, 1} I'm pretty sure!!

8 0

4 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

3 years ago

1. How many Grade 10C students should join so that they can fill the quota? (Show your computation).

expeople1 [14]

You have to put the b is

3 0

2 years ago

Zack bought a coat for $69.78. He paid with $100 bill and received $26.73 in change. How much was the sales tax?

Klio2033 [76]

$3.49

Hopefully I helped!

6 0

3 years ago

Read 2 more answers

Geometry math question

Brut [27]

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

$\dfrac{x}{13}{y}=\dfrac{19}{y+a}=\dfrac{x}{y+2a}$

Solve for y from first proportion

$\dfrac{13}{y}=\dfrac{19}{y+a}\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=\dfrac{13}{6}a$

Substitute to the second proportion

$\dfrac{19}{y+a}=\dfrac{x}{y+2a}\\\downarrow\\\dfrac{19}{\frac{13}{6}a+a}=\dfrac{x}{\frac{13}{6}a+2a}\\\\\dfrac{19}{\frac{13}{6}a+\frac{6}{6}a}=\dfrac{x}{\frac{13}{6}a+\frac{12}{6}a}\\\\\dfrac{19}{\frac{19}{6}a}=\dfrac{x}{\frac{25}{6}a}\ \ \ \ |\dot a\neq0\\\\\dfrac{19}{\frac{19}{6}}=\dfrac{x}{\frac{25}{6}}\\\\19\cdot\dfrac{6}{19}=x\cdot\dfrac{6}{25}\\\\6=\dfrac{6x}{25}\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25$

Answer: B. 25

7 0

4 years ago