<u>Given</u>:
The given circle with center at C. The lines AB and AD are tangents to the circle C.
The length of AB is (3x + 10)
The length of AD is (7x - 6)
We need to determine the value of x.
<u>Value of x:</u>
Since, we know the property of tangent that, "if two tangents from the same exterior point are tangent to a circle, then they are congruent".
We shall determine the value of x using the above property.
Thus, we have;
AB = AD
Substituting the values, we get;
Subtracting both sides of the equation by 7x, we get;
Subtracting both sides of the equation by 10, we get;
Dividing both sides of the equation by -4, we get;
Thus, the value of x is 4.
Answer:
C)
The rate of change/slope is constant, when you calculate it for this problem, it doesn't change for any amount of marbles listed in the table
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:
Honestly agreed, i feel like the smaller grades k- 6th grade are useful. where yoh learn to add, multiply, read time and such.
