Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.
Therefore, the angle BKQ is equal to 180-50-45=85°.
Of course angle BKP=180-85=95°.
Hope this helps :)
X/a=0.5
y/a=0.5
X+y/a
a=2
X=1
y=1
x+y=a(1)
1+y=2(1)
y=2-1
y=1
Answer:
2458
Step-by-step explanation:
Answer:
what the choices?
Step-by-step explanation:
Answer:
k = -2
Step-by-step explanation:
x -1 0 2 5
f(x) 2 0 -4 -10
ƒ(x) = kx
Substitute a pair of values for x and ƒ(x)
-10 = k×5
Divide each side by 5
k = -2
The constant of variation k = -2.