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3 years ago

What is the length of c in inches

Mathematics

1 answer:

crimeas [40]3 years ago

4 0

Answer:

$\sqrt{500}$ inches

Step-by-step explanation:

a^2+b^2=c^2

10^2+20^2=c^2

100+400=c^2

c^2=500

$\sqrt{500}$

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Read 2 more answers

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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

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

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