Can u help with this one u know who u are

Mathematics

2 answers:

tatiyna3 years ago

8 0

It's the one on the top righr, -2/9 w + 2/15.

liraira [26]3 years ago

8 0

1/3(2/5-2/3w)

1/3*2/5-1/3*2/3w

2/15-2/6w

2/15-1/3w

Final Answer: 2/15-1/3w

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Vertices of the triangle are: A(7, 5), B(4, 2), and C(9, 2)

According to the diagram below....

Length of the side BC (a) $=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5$

Length of the side AC (b) $= \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}$

Length of the side AB (c) $= \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}$

We need to find ∠ABC or ∠B . So using Cosine rule, we will get...

$cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree$