What is the equation of a circle whose center is at the origin and contains the point (7, 0)?

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

7 0

(x-center)^2+(y-center)^2=r^2
x^2+y^2=r^2
(7)^2+(0)^2=r^2
49+0=r^2
r^2=49, r=7
x^2+y^2=49

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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)

Gekata [30.6K]

Answer:

The pentagon MNPQR has a perimeter of 22 units.

Step-by-step explanation:

Geometrically speaking, the perimeter of the pentagon is the sum of the lengths of each side, that is:

$p = MN + NP + PQ + QR + RM$ (1)

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

If we know that $M(x,y) = (2,4)$ , $N(x,y) = (5,8)$ , $P(x,y) = (8,4)$ , $Q(x,y) = (8,1)$ and $R(x,y) = (2,1)$ , then the perimeter of the pentagon MNPQR is:

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