<h3>Answer:</h3>
<h3>Explanation:</h3>
It can work well to consider the function in parts. Define the following:
... a(x) = (1/2)ln(x^2+3)
... b(x) = x(4x^2-1)^3
Then the derivatives of these are ...
... a'(x) = (1/2)·1/(x^2 +3)·2x = x/(x^2+3)
... b'(x) = (4x^2 -1)^3 + 3x(4x^2 -1)^2·8x = (4x^2 -1)^2·(4x^2 -1 +24x^2)
... = (4x^2 -1)^2·(28x^2 -1)
___
<em>Putting the parts together</em>
f(x) = a(x)/b(x)
f'(x) = (b(x)a'(x) -a(x)b'(x))/b(x)^2 . . . . . rule for quotient of functions
Substituting values, we have
... f'(x) = (x(4x^2 -1)^3·x/(x^2 +3) -(1/2)ln(x^2 +3)·(4x^2 -1)^2·(28x^2 -1)) / (x(4x^2 -1)^3)^2
We can cancel (4x^2 -1)^2 from numerator and denominator. We can also eliminate fractions (1/2, 1/(x^2+3)). Then we have ...
... f'(x) = 2x^2(4x^2 -1) -(x^2 +3)ln(x^2 +3)·(28x^2 -1)/(2x^2·(x^2 +3)(4x^2 -1)^4))
Simplifying a bit, this becomes ...
... f'(x) = (8x^4 -2x^2 -ln(x^2 +3)·(28x^4 +83x^2 -3))/(2x^2·(x^2 +3)(4x^2 -1)^4))
To answer this problem we have to have a <span>Standard Normal Distribution table because we need to look up z scores. Then we use this equation
</span><span>z = (value - mean)/(standard deviation) = (250-200)/20 = 2.5.
</span><span>
The z score is then 2.5 </span>If so, look up z=2.5 on the left edge then you then determine from the table what the area to the left of that is.
<span>
Here are the conditions:
</span><span>p(z>2.5)=1
p(z<2.5)=.621
1-.621= .379
100-37.9= 62.1</span><span>
Given that, we should go with </span><span>.621%.</span>
I think it’s 27
explanation:
81
Step by step explaination
Answer: plus and times
Step-by-step explanation: You can do 5.89x 4 or
5.89+5.89+5.89+another 5.89 so thoes are your two answers