Solve for x. C=x+a/b
1 answer:
Since you are solving for x, you need to isolate x on one side of the equation. Start by subtracting C from both sides.
0 = x + a / b - C
Subtract x from both sides to have it on the left side of the equation.
-x = a / b - C
Make x a positive number by dividing both sides by negative one.
x = - a / b + C
A better way of writing this answer would be: .
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Answer:
Step-by-step explanation:
Begin with the standard form of a circle as a conic:
For a circle, A and C will be the exact same, and B will equal 0. If B is non-zero, the equation represents a rotation of a conic, which is reserved for college-level courses. Shortening this, then:
is good enough for us for this. Start with the first point on the circle, (2, 31) and fill in the equation above with x and y:
which simplifies down to:
Do the same with the next point on the circle, (-15, 14):
which simplifies down to:
Do the same with the last point, (33, 0):
which simplifies down to:
Now we will add (1) and (2) to get (4):
2D + 31E + F = -965
-15D + 14E + F = -421
Multiply the top equatio by -1 to get rid of the F terms:
-2D - 31E - F = 965
-15D + 14E + F = -421
which simplifies to
(4): -17D - 17E = 544
Now add (2) and (3) to get (5):
-15D + 14E + F = -421
33D + F = -1089
Multiply the bottom equation by -1 to get rid of the F terms:
-15D + 14E + F = -421
-33D - F = 1089
which simplifies to
(5): -48D + 14E = 668
Now add (4) and (5) together and eliminate the E terms:
-17D - 17E = 544
-48D + 14E = 668
In order to eliminate the E terms, multiply the top equation by 14 and the bottom equation by 17 to solve for D:
-238D - 238E = 7616
-816D + 238E = 11356
Which gives you that
D = -18
Now plug the value for D into (4) to find E:
-17(-18) - 17E = 544 and
306 - 17E = 544 and
-17E = 238 so
E = -14
Now plug the values for both D and E into (1) to find F:
2(-18) + 31(-14) + F = -965 and
-36 - 434 + F = -965 and
-470 + F = -965 so
F = -495
Now we can fill in the standard form of the conic:
but we're not done til we complete the square on both the x terms and the y terms (and I am assuming you know how to complete the square):
which simplifies to
The second choice down matches that equation, but the center they have there is not correct. The center of that circle is (9, 7) and they have it as being (7, 10) with a radius of 5. The radius is 25. The center is wrong in the equation that represents the circle as is the radius. Maybe let someone know that...
Answer:
<h2>h = 9, k = 8</h2>Step-by-step explanation:
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...![\bf \begin{cases} x=-2\implies x+2=0\implies &(x+2)=0\\ x=2i\implies x-2i=0\implies &(x-2i)=0\\ x=-2i\implies x+2i=0\implies &(x+2i)=0 \end{cases} \\\\\\ (x+2)\underline{(x-2i)(x+2i)}=0\\\\ -----------------------------\\\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ (x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0 \\\\\\ (x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0 \\\\\\ x^3+2x^2+4x+8=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D-2%5Cimplies%20x%2B2%3D0%5Cimplies%20%26%28x%2B2%29%3D0%5C%5C%0Ax%3D2i%5Cimplies%20x-2i%3D0%5Cimplies%20%26%28x-2i%29%3D0%5C%5C%0Ax%3D-2i%5Cimplies%20x%2B2i%3D0%5Cimplies%20%26%28x%2B2i%29%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Cunderline%7B%28x-2i%29%28x%2B2i%29%7D%3D0%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bdifference%20of%20squares%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%28a-b%29%28a%2Bb%29%20%3D%20a%5E2-b%5E2%5Cqquad%20%5Cqquad%20%0Aa%5E2-b%5E2%20%3D%20%28a-b%29%28a%2Bb%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%282i%29%5E2%5D%3D0%5Cimplies%20%28x%2B2%29%5Bx%5E2-%282%5E2i%5E2%29%5D%3D0%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%284%5Ccdot%20-1%29%5D%3D0%5Cimplies%20%28x%2B2%29%28x%5E2%2B4%29%3D0%0A%5C%5C%5C%5C%5C%5C%0Ax%5E3%2B2x%5E2%2B4x%2B8%3D0)
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15