6th grade
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<h2>The distance to the Eath's Horizon from point P is 352.8 mi, approximately.</h2>Step-by-step explanation:
You observe the problem from a graphical perspective with the image attached.
Notice that side is tangent to the circle, which means is perpendicular to the radius which is equal to 3,959 mi.
We have a right triangle, that means we need to use the Pythagorean's Theorem, to find the distance to the Earth's Horizon from point P.
The hypothenuse is 3959 + 15.6 = 3974.6 mi.
Therefore, the distance to the Eath's Horizon from point P is 352.8 mi, approximately.
Our function f(x) can be rewritten if we factor out a common x^2 from each term:
Now inside the parentheses we have a polynomial of the form a^2 - b^2, or the difference of two perfect squares, which can be factored as (a+b)(a-b) so we have:
Setting our function equal to zero gives us the roots x = 0, x = 4, and x = -4.
The multiplicity of the root zero is two since it occurs twice, and the others are one since they occur only once. If you graph the function you can see that it will only touch the x-axis at x = 0, but will cross the x-axis at x = 4 and x = -4.
Now inside the parentheses we have a polynomial of the form a^2 - b^2, or the difference of two perfect squares, which can be factored as (a+b)(a-b) so we have:
Setting our function equal to zero gives us the roots x = 0, x = 4, and x = -4.
The multiplicity of the root zero is two since it occurs twice, and the others are one since they occur only once. If you graph the function you can see that it will only touch the x-axis at x = 0, but will cross the x-axis at x = 4 and x = -4.