Answer:
Draw a 3-D of a terrain indicating different fault behaviors. Use the terrain below. Be sure that the entire terrain should cover at least three type of fault. Labcl the fault arcas and its part
The solution to the above factorization problem is given as f′(x)=4x³−3x²−10x−1. See steps below.
<h3>What are the steps to the above answer?</h3>
Step 1 - Take the derivative of both sides
f′(x)=d/dx(x^4−x^3−5x^2−x−6)
Step 2 - Use differentiation rule d/dx(f(x)±g(x))=d/dx(f(x))±d/dx(g(x))
f′(x)=d/dx(x4)−d/dx(x^3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−d/dx(x3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x2−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−1−dxd(6)
f′(x)=4x^3−3x^2−10x−1−0
Learn more about factorization:
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I would compute sqrt(1 + 160pi^2) first to get approximately 39.75093337
Add this to 1 and we have 40.75093337
Then divide over 2pi to get a final approximate result of 6.48571248
So x = 6.48571248 is one approximate solution
In short, I computed
only focusing on the plus for now.
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If you were to compute
you should get roughly -6.167402596 as your other solution. Each solution can then be plugged into the original equation to check if you get 0 or not. You likely won't land exactly on 0 but you'll get close enough.
Point Slope Formula: y-y₁=m(x-x₁)
Linear Standard Form: Ax+By=C
Slope Intercept: y=mx+b