To prove part of the triangle midsegment theorem using

Mathematics

2 answers:

timama [110]2 years ago

4 0

Answer:

Step-by-step explanation:

bonufazy [111]2 years ago

3 0

Answer:the length of GH is half the length of KL

Step-by-step explanation: Just took it on edg and was correct.

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Sin5A/SinA-cos5A/cosA=4cos2A

irina1246 [14]

Answer:

See Explanation

Step-by-step explanation:

$\frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ LHS = \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} \\ \\ = \frac{ \sin5A \:\cos A - \cos5A \: \sin A}{\sin A \:\cos A } \\ \\ = \frac{ \sin(5A -A )}{\sin A \:\cos A} \\ \\ = \frac{ \sin 4A}{\sin A \:\cos A} \\ \\ = \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ =4\cos 2A \\ \\ = RHS \\ \\ thus \\ \\ \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ hence \: proved$

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3 years ago

Which inequality is true for x = 2?

SOVA2 [1]

B is the true inequality

7 0

2 years ago

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Is 3, 1, 1/3, 1/9,... a geometric sequence?

barxatty [35]

$a_1;\ a_2;\ a_3;\ a_3;\ ...;\ a_n;\ a_{n+1};...\ is\ a\ geometric\ sequence\\if\ a_{n+1}:a_n=const.\\-------------------\\a_1=3;\ a_2=1;\ a_3=\dfrac{1}{3};\ a_4=\dfrac{1}{9}\\\\a_4:a_3=\dfrac{1}{9}:\dfrac{1}{3}=\dfrac{1}{9}\cdot\dfrac{3}{1}=\dfrac{1}{3}\\\\a_3:a_2=\dfrac{1}{3}:1=\dfrac{1}{3}\\\\a_2:a_1=1:3=\dfrac{1}{3}\\\\Answer:Yes.\ It's\ the\ geometric\ sequence.$