True/False - no justification required Note: each sub item is a seperate true/fa

Question

True/False - no justification required

Note: each sub item is a seperate true/false question

Consider a sector of a disk with a radius1, as in Figure, If the angle of the sector in theta in radians, Then the are along the circumference cut out by the sectoris of length theta. Sin^2x + cos^2 x = 1 is true for any real number x. Since 1/x x^-1 , we have Integrate 1/x dx = Integrate x^-1 dx = x^b/-1 + c. It is known that the function f(x) = 1/2 (x square root of x^2 + 1 In(z + square root of 1 + x^2) calculates the area under the graph of y = square root of t^2 + 1 between t = 0 and = x , Based on this fact, we can conclude that f'(x) = square root of x^2 + 1. Let f (x) = -in(cos x) for 0 is less than x is less than pie/2. Because d/dx f(x) = tan(x), we can conclude that the area under the graph of y = tan(x) between x = 0 and x = pie/4 is equal to f (pie/4) -f(0). The function f(x) = sin (x) + 1 is a primitive of the function g(x) = cos(x). delta/deltay e^x = 0

Explanation / Answer

1)yes

2)yes

3)No

4)no

5)yes

6)yes

7)yes

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