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MAXImum [283]
4 years ago
8

What is the equation of a circle with center (2, -5) and radius 4?

Mathematics
2 answers:
satela [25.4K]4 years ago
8 0
Equation\ of\ circle\\\\\
r^2=(x-a)^2+(y-b)^2\\\\
a,b- \ coordinates\ of\ center\ of\ circle\\\\\
(a,b)=(2,-5)
\\\\ r-\ radius\\\\
r=4\\\\Solution:\ 4^2=(x-2)^2+(y+5)^2\\\\\


Anna35 [415]4 years ago
4 0
(x-h)^2+(y-k)^2=r^2\\\\
(x-2)^2+(y+5)^2=16
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a) 2x2 b)3x3 c) Not possible d) 2x3 e) 3x2 f) 3x3 g)Not possible h)3x2

Step-by-step explanation:

-> we can only add matrices with same size

-> we can only multiply when columns of first equal to rows of second

-> nxm multiplied by pxq gives nxq eg 3x2 mul 2x2 gives 3x2

-> addition and mul by const number dont change the size

So considering the above facts :-

a) AB+C = [(2x2)(2X2)]+(2x2)=(2x2)+(2x2)=2x2

b) 3GF =3[(3x3)(3x3)]=3(3x3)=3x3

c)CK+B= [(2x2)(2x3)]+(2x2)=(2x3)+(2x2) which is not possible (see point 1)

d)CK+H =as CK is (2x3) so (2x3)+(2x3)=(2x3)

e)EMC=[(3x3)(3x2)](2x2)=(3x2)(2x2)=3x3

f)GLH=[(3x3)(3x2)](2x3)=(3x2)(2x3)=3x3

g)HLG=[(2x3)(3x2)](3x3)=(2x2)(3x3)= not possible(see point 2)

h)2EL+5MB=2[(3x3)(3x2)]+5[(3x2)(2x2)]=(3x2)+(3x2)=3x2

   

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