The resolvent is:
x = (- b +/- root (b2 - 4ac)) / 2a
To apply it we must have a polynomial of the form:
ax2 + bx + c = 0
Where,
One side of the equation is zero.
The polynomial must be only grade 2.
The coefficient a must be different from zero.
Answer:
options: B, C, D are correct
Answer:
Eq: (x+a/2)²+(y+1)²=(a²-8)/4
Center: O(-a/2, -1)
Radius: r=0.5×sqrt(a²-8)
Mandatory: a>2×sqrt(2)
Step-by-step explanation:
The circle with center in O(xo,yo) and radius r has the equation:
(x-xo)²+(y-yo)²=r²
We have:
x²+y²+ax+2y+3=0
But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4
And
y²+2y+3=y²+2y+1+2=(y+1)²+2
Replacing, we get:
(x+a/2)²-a²/4+(y+1)²+2=0
(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4
By visual inspection we note that:
- center of circle: O(-a/2, -1)
- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)
Answer: 2nd graph and the point is (6,7)
Step-by-step explanation:
Answer:
In order to find the value of s, we must isolate the s variable in this equation, so the first step should be subtracting 3 5/6 from both sides of this equation. In order to subtract, we multiply each denominator to get 42, and multiply the numerator by the same amount, and subtract the number from both sides:
s + 3 35/42 = 9 6/42
s = 5 13/42
Step-by-step explanation:
