Answer:
(x + 2)^2-9
How to find answer:
Use the formula (b/2)^2 in order to complete the square.
Hope this helps :)
X+0=18
2x+y=19
(-2)x+(-2)0=(-2)18
2x+y=19
0+y= -17
2x+ (-17)=19
2x=36
y=-17
x=18
All right angle triangles have perpendicular lines
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
• 1 + cot² x = csc²x and csc x = ![\frac{1}{sinx}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bsinx%7D)
• sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x
Consider the left side
sin²Θ( 1 + cot²Θ )
= sin²Θ × csc²Θ
= sin²Θ × 1 / sin²Θ = 1 = right side ⇔ verified
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Consider the left side
cos²Θ - sin²Θ
= cos²Θ - (1 - cos²Θ)
= cos²Θ - 1 + cos²Θ
= 2cos²Θ - 1 = right side ⇒ verified
Answer:
![(x+4)^2+(y-3)^2=149](https://tex.z-dn.net/?f=%28x%2B4%29%5E2%2B%28y-3%29%5E2%3D149)
Step-by-step explanation:
The equation of a circle is
where
is the center of the circle and
is the radius of the circle.
Given that
and it passes
, their distance between each other must the radius of the circle, so we can use the distance formula to find the radius:
![d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\d=\sqrt{(-4-3)^2+(6-(-4))^2}\\\\d=\sqrt{(-7)^2+10^2}\\\\d=\sqrt{49+100}\\\\d=\sqrt{149}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y_2-y_1%29%5E2%2B%28x_2-x_1%29%5E2%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B%28-4-3%29%5E2%2B%286-%28-4%29%29%5E2%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B%28-7%29%5E2%2B10%5E2%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B49%2B100%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B149%7D)
Therefore, if the length of the radius is
units, then
, making the final equation of the circle ![(x+4)^2+(y-3)^2=149](https://tex.z-dn.net/?f=%28x%2B4%29%5E2%2B%28y-3%29%5E2%3D149)