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seropon [69]
3 years ago
9

Look at the equation shown below.

Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

<em>Hey</em><em>!</em><em>!</em>

<em>Sol</em><em>ution</em><em>,</em>

<em>-6x-8</em><em>=</em><em>4</em>

<em>or</em><em>,</em><em> </em><em>-6x</em><em>=</em><em>4</em><em>+</em><em>8</em>

<em>or</em><em>,</em><em> </em><em>-6x</em><em>=</em><em>1</em><em>2</em>

<em>or</em><em>,</em><em>X=</em><em>1</em><em>2</em><em>/</em><em>-</em><em>6</em>

<em>X=</em><em>-</em><em>2</em>

<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>-</em><em>2</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em> </em><em>helps</em><em>.</em><em>.</em>

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

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6 0
3 years ago
Select all the correct answers. Which two inequalities can be used to find the solution to this absolute value inequality? 3|x +
ziro4ka [17]

The absolute value inequality can be decomposed into two simpler ones.

x < 0

x > -8

<h3></h3><h3>Which two inequalities can be used?</h3>

Here we start with the inequality:

3|x + 4| - 5 < 7

First we need to isolate the absolute value part:

3|x + 4| < 7 + 5

|x + 4| < (7 + 5)/3

|x + 4| < 12/3

|x + 4| < 4

The absolute value inequality can now be decomposed into two simpler ones:

x + 4 < 4

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Solving both of these we get:

x < 4 - 4

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These are the two inequalities.

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brainly.com/question/24372553

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6 0
9 months ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
Line segment AB is shown on a coordinate grid:
Lady_Fox [76]

Answer:

a'b' and ab are the same length

Step-by-step explanation:

they are both four squares long

7 0
2 years ago
What is -6 -8 -10 -5 -2 -3 -1 from least to greatest?
WINSTONCH [101]

Answer:

-10, -8, -6, -5, -2, -1

Step-by-step explanation:

I knew this because every negative number can still relate to being the least number. But hey, hope this helps. :]

8 0
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