Answer:
i think it A
Step-by-step explanation:
Answer:
Step-by-step explanation:
The answer is 1092m
Answer:
y=-3
Step-by-step explanation:according to the graph, for all values of x, y is equal to -3, so this function is independent of the variable x, so the equation is y=-3 (constant for all values of the Real numbers)
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)
The given equation has no solution when K is any real number and k>12
We have given that
3x^2−4x+k=0
△=b^2−4ac=k^2−4(3)(12)=k^2−144.
<h3>What is the condition for a solution?</h3>
If Δ=0, it has 1 real solution,
Δ<0 it has no real solution,
Δ>0 it has 2 real solutions.
We get,
Δ=k^2−144 here Δ is not zero.
It is either >0 or <0
Δ<0 it has no real solution,
Therefore the given equation has no solution when K is any real number.
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