How do you find the squared foot of an octagon?
1 answer:
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Given :
The square of a number, x, is 16 less than eight times the number.
To Find :
The value of x .
Solution :
The mathematical equation with all given data is :
Therefore , the value of x is 4 .
Hence , this is the required solution .
Circle formula
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius
to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=
points (-3,2) and (1,5)
D=
D=
D=
D=
D=5
center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1
the point (-7,5) and (7,-1) lie on this circle
radius=5 units
the points (-7,5) and (-7,1) lie on this circle
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius
to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=
points (-3,2) and (1,5)
D=
D=
D=
D=
D=5
center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1
the point (-7,5) and (7,-1) lie on this circle
radius=5 units
the points (-7,5) and (-7,1) lie on this circle