The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
<h3>What is the derivative of the function g(x) by virtue of the Fundamental theorem of calculus as given in the task content?</h3>
g(x) = Integral; √2 ln(t) dt (with the upper and lower limits e^x and 1 respectively).
Since, it follows from the Fundamental theorem of calculus that given an integral where;
Now, g(x) = Integral f(t) dt with limits a and x, it follows that the differential of g(x);
g'(x) = f(x).
Consequently, the function g'(x) which is to be evaluated in this scenario can be determined as:
g'(x) = 
= 1
The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
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i think it is A. P(5<X<9)
5.8
a2+b2=c and the once you have c which was
34 you find the square root of 34 which is 5.8 i’m kind of positive that i’m right
Answer:
see the explanation
Step-by-step explanation:
we know that
If the absolute value of the scale factor is less than 1, then the dilation produces a contraction of the original image
If the absolute value of the scale factor is greater than 1, then the dilation produces an expansion of the original image
so
<u><em>Verify each value</em></u>
1) -4
therefore
The dilation produces an expansion of the original image
2) 0.25
therefore
The dilation produces a contraction of the original image
3) -2/3
therefore
The dilation produces a contraction of the original image
4) 2.3
therefore
The dilation produces an expansion of the original image
Answer:
Grouping and then use the common factor. Therefor the answer is 6(2p+5)
Step-by-step explanation:
I hope this helps.
