We know: The sum of the measures of the angles of a triangle is equal 180°.
We have: m∠A =65°, m∠B = (3x - 10)° and m∠C = (2x)°.
The equation:
65 + (3x - 10) + 2x = 180
(3x + 2x) + (65 - 10) = 180
5x + 55 = 180 <em>subtract 55 from both sides</em>
5x = 125 <em>divide both sides by 5</em>
x = 25
m∠B = (3x - 10)° → m∠B = (3 · 25 - 10)° = (75 - 10)° = 65°
m∠C = (2x)° → m∠C = (2 · 25)° = 50°
<h3>Answer: x = 25, m∠B = 65°, m∠C = 50°</h3>
The answer to this question is 36
From what I'm reading from this question, there's one answer.
13 + 5 * 1 = 18
Unless maybe I'm missing something?
They are always on a coplanar
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
__
For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
__
For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
___
If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
_____
Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.