What power can you raise both sides of (2x - 1) = xź to transform the fractional exponents to

Mathematics

2 answers:

liubo4ka [24]3 years ago

5 0

2x1=2-1=1 hope this helps

SSSSS [86.1K]3 years ago

5 0

3 is the right answer

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Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$