Interesting problem ...
The key is to realize that the wires have some distance to the ground, that does not change.
The pole does change. But the vertical height of the pole plus the distance from the pole to the wires is the distance ground to the wires all the time. In other words, for any angle one has:
D = L * sin(alpha) + d, where D is the distance wires-ground, L is the length of the pole, alpha is the angle, and 'd' is the distance from the top of the (inclined) pole to the wires:
L*sin(40) + 8 = L*sin(60) + 2, so one can get the length of the pole:
L = (8-2)/(sin(60) - sin(40)) = 6/0.2232 = 26.88 ft (be careful to have the calculator in degrees not rad)
So the pole is 26.88 ft long!
If the wires are higher than 26.88 ft, no problem. if they are below, the concerns are justified and it won't pass!
Your statement does not mention the distance between the wires and the ground. Do you have it?
Answer:
is there a picture i dont see them
Step-by-step explanation:
5.6 is the correct answer because 1.3 +4.3 would be 5.6 so the next one in the sequence is 9.9
To figure out how much she spent all we have to do is add up $0.85, $4.50 and $1.50
$0.85 + $4.50 + $1.50 = $6.85
Therefore she spent $6.85
Hope this helps
Answer:
The coordinates of H′ are (-2,0).
Step-by-step explanation:
It is given that square EFGH stretches vertically by a factor of 2.5 with respect to the x-axis to create rectangle E′F′G′H′.
If a figure stretches vertically by a factor of 2.5 with respect to the x-axis, then the x-coordinates remains same and the points which lie on the axis are also remains the same after stretch.
It is given that the coordinates of H are (-2,0). This point lie on the x-axis, therefore it will remains the same after stretch.
Therefore the coordinates of H′ are (-2,0).
