F(3) = x + 2; in this equation what is the argument?
1 answer:
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Point P = base of the pole
Point Q = point where the pole broke
Point R = point where the tip of the pole landed on the ground.
We're told that "28 feet are still sticking straight out of the ground", so PQ = 28 is the vertical leg of the right triangle. The horizontal leg is unknown, so we'll call it x, making PR = x. The hypotenuse is QR = 53 because 81-28 = 53. This is the left-over amount after taking off the 28 feet initially out of the ground from the 81 ft pole.
In short, we have these three values
a = x is one of the legs
b = 28 is the other leg
c = 53 is the hypotenuse
Use the Pythagorean Theorem to find x
a^2 + b^2 = c^2
x^2 + 28^2 = 53^2
x^2 + 784 = 2809
x^2 = 2809 - 784
x^2 = 2025
x = sqrt(2025)
x = 45
The end of the pole is 45 feet away from the base
Answer: 45 feet
Point P = base of the pole
Point Q = point where the pole broke
Point R = point where the tip of the pole landed on the ground.
We're told that "28 feet are still sticking straight out of the ground", so PQ = 28 is the vertical leg of the right triangle. The horizontal leg is unknown, so we'll call it x, making PR = x. The hypotenuse is QR = 53 because 81-28 = 53. This is the left-over amount after taking off the 28 feet initially out of the ground from the 81 ft pole.
In short, we have these three values
a = x is one of the legs
b = 28 is the other leg
c = 53 is the hypotenuse
Use the Pythagorean Theorem to find x
a^2 + b^2 = c^2
x^2 + 28^2 = 53^2
x^2 + 784 = 2809
x^2 = 2809 - 784
x^2 = 2025
x = sqrt(2025)
x = 45
The end of the pole is 45 feet away from the base
Answer: 45 feet