I need help with this
1 answer:
Answer:
Step-by-step explanation:
Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.
__
Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².
__
Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).
__
Taken altogether, these factors give us the function ...
You might be interested in
Answer:
Exercise (a)
The work done in pulling the rope to the top of the building is 750 lb·ft
Exercise (b)
The work done in pulling half the rope to the top of the building is 562.5 lb·ft
Step-by-step explanation:
Exercise (a)
The given parameters of the rope are;
The length of the rope = 50 ft.
The weight of the rope = 0.6 lb/ft.
The height of the building = 120 ft.
We have;
The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;
ΔW₁ = 0.6Δx·x
The work done for the second half, ΔW₂, is given as follows;
ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375
The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375
∴ We have;
W =
The work done in pulling the rope to the top of the building, W = 750 lb·ft
Exercise (b)
The work done in pulling half the rope is given by W₂ as follows;
The work done in pulling half the rope, W₂ = 562.5 lb·ft