Sagot :
approx. 144 mileswe can find this length by use of pythagorean's threorom. we use it because there's a 90 degree angle in the triangle, first we go due north, then turn due west.
[tex] a^{2} [/tex] + [tex] b^{2} [/tex] = [tex] c^{2} [/tex] , where a & b and the lines attached to the right angle, and c is the hypotenuse, also the length we're looking for.
[tex] 120^{2} [/tex] + [tex] 80^{2} [/tex] = [tex] c^{2} [/tex]
14400 + 6400 = [tex] c^{2} [/tex]
20800 = [tex] c^{2} [/tex]
c = [tex] \sqrt{20800} [/tex] ≈ 144.2
[tex] a^{2} [/tex] + [tex] b^{2} [/tex] = [tex] c^{2} [/tex] , where a & b and the lines attached to the right angle, and c is the hypotenuse, also the length we're looking for.
[tex] 120^{2} [/tex] + [tex] 80^{2} [/tex] = [tex] c^{2} [/tex]
14400 + 6400 = [tex] c^{2} [/tex]
20800 = [tex] c^{2} [/tex]
c = [tex] \sqrt{20800} [/tex] ≈ 144.2
Approximately 144 miles. You have to use the pythagorean theorem. a^2+b^2=c^2.
120 and 80 will be the legs, a and b. 120^2 + 80^2 = c^2. 14400 + 6400 = 20800. Remember, c is squared, so to find the answer, you need to find the square root of 20800, which is 144.222051019. Therefore, c, the road between Hermansville and Melville, would be approximately 144 miles.
120 and 80 will be the legs, a and b. 120^2 + 80^2 = c^2. 14400 + 6400 = 20800. Remember, c is squared, so to find the answer, you need to find the square root of 20800, which is 144.222051019. Therefore, c, the road between Hermansville and Melville, would be approximately 144 miles.