<span>∫[(secx)−1<span>]<span>(1/2)</span></span>dx?</span> or <span>∫[sec(x−1)<span>]<span>(1/2)</span></span>dx hope it works</span>
If the octopus was 3 feet BELOW the surface than that means it was at -3 but if the octopus swam up 3 feet that means the octopus location is at 0 now.
So that means the equation would look like this -3+3=0
Scenario:
a worker needs to repair a window on the second floor of a building
He is outside 3 feet way from the building
He starts walking and when he reaches the wall he is lifted 4 feet until reaching the window
What is the distance from the window to the point where he has standing before he start walking ?
2))This path forms a right triangle.2 sides are known but not the hypotenuse
Formula a^=b^=c^
a=4 b=3 c=x (Hypotenuse)
3)
Substitute
4^+3^=X^
16+9=x^
25=x^
5=x
Hypotenuse=5
The sides of the triangle formed in this situation are
4,3,and 5
