Answer:
(p, q) = (8, 82)
Step-by-step explanation:
When a circle is centered at the origin, the radius to point (a, b) will have slope m = b/a. The tangent is perpendicular to the radius, so the tangent at point (a, b) will have slope -a/b. In point-slope form, the equation of the tangent line will be ...
y -k = m(x -h) . . . . . point-slope equation of line with slope m through (h, k)
y -b = (-a/b)(x -a)
Rearranging this to standard form, we have ...
b(y -b) = -a(x -a)
by -b² = -ax +a²
ax +by = a² +b²
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For (a, b) = (5, 4), the standard form equation of the tangent can be written ...
5x +4y = 5² +4² = 41
Your given equation has an x-coefficient that is twice the value shown in this equation, so we need to multiply this equation by 2:
2(5x +4y) = 2(41)
10x +8y = 82
Comparing to 10x +py = q, we see that ...
p = 8
q = 82
I believe the answer is x^2-4x+1
(I did the problem in my head but it seems correct)
Answer:
17.5
Step-by-step explanation:
Answer:
There is no solution
Step-by-step explanation:
4x - 2y = 5 --> 4x - 2(2x + 10) = 5
4x - 4x - 20 = 5
4x - 4x -25 = 0
Final answer is -25 = 0 and so therefor it is no solution
Answer:
Step-by-step explanation:
Let 
Subbing in:
a = 9, b = -2, c = -7
The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:
Group together in groups of 2:
Now factor out what's common within each set of parenthesis:
We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:
Remember that 
so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.
The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring 
gives us that x = 1 and -1. The other set is a bit more tricky. If
then
and
You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:
±
which will simplify down to
±
Those are the 4 solutions to the quartic equation.
