Answer:
(g, f) = (-1, 2)
Step-by-step explanation:
The equation can be put into standard form:
(x^2 +2gx) +(y^2 +2fy) = 15
(x^2 +2gx +g^2) +(y^2 +2fy +f^2) = 15 +g^2 +f^2
(x +g)^2 +(y +f)^2 = 15 +g^2 +f^2 . . . . equation in standard form
The point (3, 2) is on the circle, so one of the equations we have is ...
(3+g)^2 +(2 +f)^2 = 15 +g^2 +f^2
9 +6g +g^2 +4 +4f +f^2 = 15 +g^2 +f^2
6g +4f = 2 . . . . . . . subtract g^2 +f^2 +13
3g +2f = 1 . . . . . . . divide by 2
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A line through the center of the circle is perpendicular to the tangent at the point of tangency. That line will have a slope that is the negative reciprocal of the tangent's slope: m = -1/(-1/2) = 2. Then in point-slope form the equation of the line through the center is ...
y -2 = 2(x -3)
2x -y = 4 . . . . rearrange to standard form
The point at the center of the circle, (-g, -f), satisfies this equation, so we have another relation for f and g:
2(-g) -(-f) = 4
2g -f = -4 . . . . . . multiply by -1
Adding twice this equation to the one above, we have ...
2(2g -f) +(3g +2f) = 2(-4) +(1)
7g = -7 . . . . . . . . . simplify
g = -1 . . . . . . . . . . divide by 7
f = 2g +4 = 2(-1) +4 = 2 . . . find the corresponding value of f
The values of f and g are: f = 2 and g = -1.
Okay, here we have this:
Considering the provided information and grades, we are going to calculate the requested GPA, so we obtain the following:
Then we will substitute in the following formula:
Then:
Finally we obtain that the student's GPA for that term is approximately 3.06.
we have
Divide both sides by
Square root both sides
so
the solutions are
therefore
<u>the answer is the option</u>
x = 2/3 and x = -2/3