Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
N = the number
0.5n - 9 = 4n + 5
(0.5n - 0.5n) - 9 = (4n - 0.5n) + 5
-9 = 3.5n + 5
(-9 - 5) = 3.5n + (5 - 5)
-14 = 3.5n
-14/3.5 = 3.5n/3.5
-4 = n
The number is -4.
The answer is C) y = 4x - 3
B and D can be eliminated by substituting 0 for x in each...the result is not -3
Now you have a 50/50 chance of getting the question correct
Answer:
-13x^2 + 6x + 11
Step-by-step explanation:
Just subtract the numbers that are similar (i.e. -5x^2 - 8x^2)
