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Yuri [45]
3 years ago
13

Solve the absolute value inequality 5|3-2x|+1<4

Mathematics
1 answer:
Gekata [30.6K]3 years ago
5 0
5\cdot|3-2x|+1\ \textless \ 4\qquad|-1\\\\5\cdot|3-2x|+1-1\ \textless \ 4-1\\\\5\cdot|3-2x|\ \textless \ 3\qquad|:5\\\\|3-2x|\ \textless \ \dfrac{3}{5}\\\\\\\\-\dfrac{3}{5}\ \textless \ 3-2x\ \textless \ \dfrac{3}{5}\qquad\Big|-3\\\\\\&#10;-\dfrac{3}{5}-3\ \textless \ 3-2x-3\ \textless \ \dfrac{3}{5}-3\\\\\\-\dfrac{18}{5}\ \textless \ -2x\ \textless \ -\dfrac{12}{5}\qquad\Big|:(-2)\\\\\\\dfrac{9}{5}\ \textgreater \ x\ \textgreater \ \dfrac{6}{5}\\\\\\\boxed{x\in\left(\frac{6}{5};\frac{9}{5}\right)}

or:

\boxed{x\in\left(1.2;1.8\right)}

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4 years ago
There is a circle with a center of 0,0 on a coordinate plane. There is one point on the circle's circumference in which the x:y
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Step-by-step explanation:

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8 0
3 years ago
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7 0
3 years ago
Simplify please. Picture is attached
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3 0
3 years ago
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Marizza181 [45]

Answer:

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Step-by-step explanation:

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