The line of sight to the horizon , the radius at the horizon and the radius at the mountain from a right angle triangle, so Pythagoras's theorem gives d = sqrt( ((R+h)^2 - R^2) = sqrt( 2Rh + h^2)
= 146.2mi
the answer is 146.2
do we need to find the point at the centre of the earth or the point on the circumfernce of earth?
nevermind
yes me please
no problem
? * 3 + 4 ÷ 4 = ? - 3
16 * 3 + 4 ÷ 4
48 + 4 ÷ 4
52 ÷ 4
13
13 3 less than 16
a tangent and radius meet at 90 degrees
use Pythagorean theorem
c= 3959mi + 4.5mi
c= 3963.5mi
a^2 = c^2 - b^2
a^2 = 3963.5^2 - 3959^2
a^2 = 35651.25
a= √35651.25
a= 188.8mi (nearest tenth)
the graph shifts 7 units left and 2 units up.
f(x)=x^2-9/x-3
f(x)=(x+3)(x-3)/(x-3)
x-3=0
x=3
3^2-9/3-3
9-9/3-3
0/0=0
y=0
hole at (3, 0)