Answer:
Given an angle formed by two lines with a common vertex, this page shows how to ... The above animation is available as a printable step-by-step instruction sheet, which can be ... This construction works by creating two congruent triangles
Step-by-step explanation:
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:
-3, 1, 5, 9
Step-by-step explanation:
2(2) - 7 = -3
2(4) - 7 = 1
2(6) - 7 = 5
2(8) - 7 = 9
Step-by-step explanation:
Let us consider the task to find the angle between vectors ES and EJ (the first letters are taken to name the vectors).
\overrightarrow{ES} = (4;4) - (4; -3) = \overrightarrow{(0; 7)}
ES
=(4;4)−(4;−3)=
(0;7)
\overrightarrow{EJ} = (-5; -4) - (4; -3) = \overrightarrow{(-9; -1)}
EJ
=(−5;−4)−(4;−3)=
(−9;−1)
cos \alpha=\frac{\overrightarrow{ES}*\overrightarrow{EJ}}{|\overrightarrow{EJ}|*|\overrightarrow{ES}|}cosα=
∣
EJ
∣∗∣
ES
∣
ES
∗
EJ
cos(a) = (0*(-9)+7*(-1)) / (7*9.055) = -0.11043;
a = 96,34°
Solution: 96 degrees.
