The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)
<h3>How to determine the distance between the two anchor points of the guy wire</h3>
The problem will be solved using SOH CAH TOA
let the distance between the 150 ft long guy wire and the radio antenna be x
let the distance between the 180 ft long guy wire and the radio antenna be y
cos 65° = x / 150
x = cos 65° * 150
x = 63.39 ft
The height of the antenna
sin 65° = height of antenna / 150
height of antenna = sin 65 * 150
height of antenna = 135.95
using Pythagoras theorem
(length of guy wire)² = (height of the antenna)² + (anchor distance)²
(anchor distance)² = 180² - 135.95²
anchor distance = √(180² - 135.95²)
anchor distance = 117.97
The anchor points distance apart
= 63.39 + 117.97
= 181 (to the nearest foot)
Learn more on Pythagoras theorem here:
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9514 1404 393
Answer:
5/8
Step-by-step explanation:
The area of the smaller circles is proportional to the square of the ratio of their diameters. The two smallest circles have diameters equal to 1/4 the diameter of the largest circle. Hence their areas are (1/4)^2 = 1/16 of that of the largest circle.
Similarly, the medium circle has a diameter half that of the largest circle, so its area is (1/2)^2 = 1/4 of the are of the largest circle.
The smaller circles subtract 2×1/16 +1/4 = 3/8 of the area of the largest circle. Then the shading is 1-3/8 = 5/8 of the area of the largest circle.
Answer:
<h3>r =-12</h3>
Step-by-step explanation:
Answer:
Step-by-step explanation:
In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:
a = -32 ft/s/s
v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)
Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)
t = ?? sec.
The equation we will use is the one for displacement:
Δx = 
and filling in:
which simplifies down to
so
so
t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)
Now we use that time in the x-dimension. Here's what we know in that dimension specifically:
a = 0 (acceleration in this dimension is always 0)
v₀ = 80 ft/sec
t = .968 sec
Δx = ?? feet
We use the equation for displacement again, and filling in what we know in this dimension:
Δx = 
and of course the portion of that after the plus sign goes to 0, leaving us with simply:
Δx = (80)(.968)
Δx = 77.46 feet
