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Kruka [31]
3 years ago
7

ANSWER THIS FAST BRANLIEST IS ON THE LINE in

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

Option A.) is correct  \triangle XYZ \sim \triangle WVZ  by AA similarity. There are two angles that are congruent given by \angle ZXY = \angle ZWV \hspace{0.2cm}  and\hspace{0.2cm}   \angle ZYX = \angle ZVW.

This is because both pairs represent alternate interior angles formed by lines intersecting two parallel lines.

Step-by-step explanation:

From the given figure we can say that

i) Option A.) is correct  \triangle XYZ \sim \triangle WVZ  by AA similarity. There are two angles that are congruent given by \angle ZXY = \angle ZWV \hspace{0.2cm}  and\hspace{0.2cm}   \angle ZYX = \angle ZVW.

This is because both pairs represent alternate interior angles formed by lines intersecting two parallel lines.

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3 years ago
Solve the inequality 2x&gt;30+5/4x
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Answer:

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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

8 0
2 years ago
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horrorfan [7]

Answer:

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Correct choice is A

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Read 2 more answers
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