(x-4)² + (x-2)² = x²
(x² - 8x + 16 ) + (x² - 4x + 4) = x
2x² -12x + 20 = x
2x²-12x + 20 - x = 0
2x² - 13x + 20 = 0
(2x-5) (x-4) = 0
2x-5 = 0
2x = 5
x = ⁵/₂
x-4 = 0
x = 4
so x₁ = ⁵/₂ , and x₂ = 4
Answer:
A) Suppose we have an onedimensional situation.
in the 0 of our x-axis, we have a fruit tree, and we want to rest at a distance no bigger than 6 ft of the tree, then all the possible positions of our resting place are:
x ∈(-6ft, 6ft)
we can write this as: IxI < 6ft
b) now we think the opposite situation, we want to rest at least 6ft away from the tree, then we have that:
x ∉ [-6ft, 6ft].
or IxI > 6ft.
So you can see that the difference in those two cases is if we want to be "inside a given range" (for the first case) or "outside a given range" (for the second case).
Average= f(2)−f(0)/2−0
=62.5−250/ 2-0
<span>= −93.75</span>
Answer:
ez no picture!!!!
Step-by-step explanation:
