Answer:
∠x = 90°
∠y = 58°
∠z = 32°
Step-by-step explanation:
he dimensions of the angles given are;
∠B = 32°
Whereby ΔABC is a right-angled triangle, and the square fits at angle A, we have;
∠A = 90°
∠B + ∠C = 90° which gives
32° + ∠C = 90°
∠C = 58°
∠x + Interior angle of the square = 180° (Sum of angles on a straight line)
∠x + 90° = 180°
∠x = 90°
∠x + ∠y + 32° = 180° (Sum of angles in a triangle)
90° + ∠y + 32° = 180°
∠y = 180 - 90° - 32° = 58°
∠y + ∠z + Interior angle of the square = 180° (Sum of angles on a straight line)
58° + ∠z +90° = 180°
∴ ∠z = 32°
∠x = 90°
∠y = 58°
∠z = 32°
Answer:
(a) draw the graph using these coordinates :
(0,-1) and (3,1)
(b) x = 3
Y = -0.01x^2 + 0.5x + 3
when the height y = 8
-0.01x^2 + 0.5x + 3 = 8
-0.01x^2 + 0.5x - 5 = 0
x = 13.82 and x = 36.18 so the ball is at least 8 ft high from 13.82 and 36.18 feet inclusive.- from the player.
If the player is 30 ft from the net then the ball is likely to go over the net.
Answer:
r = 11/2
Step-by-step explanation:
m = slope
slope, m = (y2 - y1)/(x2 - x1)
-4 = (7 - 9)/(r - 5)
cross multiply
-4(r - 5) = 7 - 9
clearing brackets
-4r + 20 = -2
-4r = -22
r = 22/4 = 11/2
