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Zigmanuir [339]
3 years ago
6

Math question please help if you get this right I will mark you as a brainliest

Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

The Law of Cosines would be your best bet here, since the unknown side is opposite the known angle 110 degrees.

|AC|^2 = (8 in)^2 + (23 in)^2 - 2(8 in)(23 in)*cos 110 degrees

= 64 in^2 + 529 in^2 - (368 in^2)*(-0.342)

= 593 in^2 + 126 in ^2 approximately

= 719 in^2 approximately

Then the length of side AC is approx. √(716 in^2) = 27 in

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Alexxx [7]
Let the marks of the students in class 1 be

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\frac{m_1+ m_2+ .....+ m_1_2}{12}=90

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m_1+ m_2+ .....+ m_1_2=90*12=1080


similarly, let n_1, n_2, ....., n_2_0 be the marks of the students in the second class, 

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which means n_1+ n_2+ ..... n_2_0=80*20=1600


The 32 students averaged:

\frac{(m_1+ m_2+ .....+ m_1_2)+(n_1+ n_2+ ..... n_2_0)}{12+20}= \frac{1080+1600}{32}= \frac{2680}{32}=   83.75
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4 years ago
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