Need help with angle relationships worksheet

Mathematics

1 answer:

Ludmilka [50]2 years ago

6 0

The information about the angles shows that:

Complementary to 40° = 50°
Complementary to 67° = 23°
Supplementary to 80° = 100°
Supplementary to 65° = 115°

<h3>What are complementary angles?</h3>

It should be noted that complementary angles refers to the angles that are equal to 90°. On the other hand, supplementary angles are the angles that are equal to 180°.

The complementary to 40°:

= 90° - 40°

= 50°

The complementary to 67° will be:

= 90° - 67°

= 23°

The supplementary to 80° will be:

= 180° - 80°

= 100°

The supplementary to 65° will be:

= 180° - 65°

= 115°

Learn more about complementary angles on:

brainly.com/question/98924

#SPJ1

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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

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