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
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Ok then after that what’s the question ?
The exponential equation of the model is A(t) = 2583 * 0.88^t and the multiplier means that the number of new cases in a week is 88% of the previous week
<h3>The function that models the data</h3>
The given parameters are:
New, A(t) = 2000
Rate, r = 12%
The function is represented as:
A(t) = A * (1 - r)^t
So, we have:
2000 = A * (1 - 12%)^t
This gives
2000 = A * (0.88)^t
2 weeks ago implies that;
t = 2
So, we have:
2000 = A * 0.88^2
Evaluate
2000 = A * 0.7744
Divide by 0.7744
A = 2583
Substitute A = 2583 in A(t) = A * 0.88^t
A(t) = 2583 * 0.88^t
Hence, the exponential equation of the model is A(t) = 2583 * 0.88^t
<h3>The interpretation of the multiplier</h3>
In this case, the multiplier is 88% or 0.88
This means that the number of new cases in a week is 88% of the previous week
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It has to be 5 because 2.5 is half so the other half 2.5 .so 2.5 +2.5 =5