What is 2*6+6/777= plz help

Mathematics

1 answer:

Nana76 [90]3 years ago

6 0

AnsStep-by-step explanation:

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

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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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2 years ago

Is the sum of two rational numbers always rational

Pepsi [2]

Answer:

in short, Yes they are always rational.

here's why...

E.g. suppose $\frac{c}{d}$ and $\frac{a}{b}$ are fractions, that means that a,b,c,d are all integers, and b and d are not zero. finding the sum the numerator, and denominator would also have to be integers. the denominator of the sum can't be zero since the denominators of the fractions were not zero, and would give $\frac{ad+bc}{bd}$ and since they are bound by addition (sum means addition) they must also be rational since eit would equal a bigger integer than initially had