Calculate the value of the exterior angle

Mathematics

1 answer:

marishachu [46]3 years ago

6 0

An exterior angle of the triangle equals to the sum opposite interior angles, so

8x+15 = (4x+5) + (3x+20)

8x +15 = 7x +25

x=10

Now, we can calculate the exterior angle.

8x+15 = 8*10 + 15 = 80+15 = 95

Answer is 95⁰.

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Find the perimeter of the pentagon MNPQR with vertices M(2, 4), N(5, 8), P(8, 4), Q(8, 1), and R(2, 1)

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$p = \sqrt{\overrightarrow{MN}\,\bullet \, \overrightarrow{MN}} + \sqrt{\overrightarrow{NP}\,\bullet \, \overrightarrow{NP}} + \sqrt{\overrightarrow{PQ}\,\bullet \, \overrightarrow{PQ}} + \sqrt{\overrightarrow{QR}\,\bullet \, \overrightarrow{QR}} + \sqrt{\overrightarrow{RM}\,\bullet \, \overrightarrow{RM}}$ (1b)

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$p =\sqrt{(5-2)^{2}+(8-4)^{2}} + \sqrt{(8-5)^{2}+(4-8)^{2}}+\sqrt{(8-8)^{2}+(1-4)^{2}}+\sqrt{(2-8)^{2}+(1-1)^{2}}+\sqrt{(2-2)^{2}+(4-1)^{2}}$ $p = \sqrt{3^{2}+4^{2}} + \sqrt{3^{2}+(-4)^{2}}+\sqrt{0^{2}+(-3)^{2}}+\sqrt{(-6)^{2}+0^{2}}+\sqrt{0^{2}+3^{2}}$