Answer:
x^2 + y^2 + 16x + 6y + 9 = 0
Step-by-step explanation:
Using the formula for equation of a circle
(x - a)^2 + (y + b)^2 = r^2
(a, b) - the center
r - radius of the circle
Inserting the values given in the question
(-8,3) and r = 8
a - -8
b - 3
r - 8
[ x -(-8)]^2 + (y+3)^2 = 8^2
(x + 8)^2 + (y + 3)^2 = 8^2
Solving the brackets
( x + 8)(x + 8) + (y +3)(y+3) = 64
x^2 + 16x + 64 + y^2 + 6y + 9 = 64
Rearranging algebrally,.
x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0
Bringing in 64, thereby changing the + sign to -
Therefore, the equation of the circle =
x^2 + y^2 + 16x + 6y + 9 = 0
1/3 oranges were spoiled.....u threw away 4 oranges...
1/3 of what = 4....
1/3x = 4
x = 4 * 3
x = 12 oranges
1/4 grapefruits were spoiled....u threw away 7
1/4 of what = 7
1/4x = 7
x = 7 * 4
x = 28
so ur fruit basket contained 12 oranges and 28 grapefruits before u discarded the spoiled ones
Answer:
B
Step-by-step explanation:
Answer:
(x, y) = (0, 1/2) or (1, 3)
Step-by-step explanation:
The first equation factors as ...
x(3x -y) = 0
This has solutions x=0 and y=3x.
__
<u>x = 0</u>
Using this in the second equation gives ...
2y -0 = 1
y = 1/2
(x, y) = (0, 1/2) is a solution
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<u>y = 3x</u>
Using the expression for y in the second equation, we get ...
2(3x) -5x = 1
x = 1 . . . . . . . . . simplify
y = 3x = 3 . . . . using x=1 in the first equation
(x, y) = (1, 3) is a solution
_____
Interestingly, the (red line) graph of 3x^2 -xy = 0 produced by this graphing calculator has a "hole" at x=0, It says that point is (0, undefined). In a sense, y is undefined, in that it can be <em>anything</em>. A more appropriate graph would graph that equation as the two lines x=0 and y=3x.