The derivative follows from the fundamental theorem of calculus:

where
is any constant in the domain of
.
We have

so

(applying the property
)
We know the bottom triangle is a 45, 45, 90 triangle, so the hypotenuse is √2 times the value of the legs:
(√2)(√2)
=√4
=2
Now, we can use this to solve for y. The top triangle is a 30, 60, 90 triangle. The side we found above is the side across from the 30 degree angle. The side opposite the 60 degree angle is √3 times the side across the 30 degree angle. Therefore, we can solve for y by multiplying 2 by √3
y=2√3
9514 1404 393
Answer:
- x=2 (work is correct)
- x=4
- x=44
- x=4
- x=any number
- impossible; no such number
Step-by-step explanation:
1. The equation is ...
3x +5 = 11
3x = 11 -5 = 6
x = 6/3 = 2 . . . . matches the work shown
__
2. The equation is ...
(x +2)·5 = 30
x +2 = 30/5 = 6
x = 6 -2 = 4
__
3. The equation is ...
(x/2) +2 = 24 . . . . matches the work shown
x/2 = 24 -2 = 22
x = 22·2 = 44
__
4. The equation is ...
2x +8 = x +12
2x = x + 12 -8 = x +4
2x -x = x -x +4
x = 4
__
5. The equation is ...
2x +4 -x -3 = x +1
x +1 = x +1 . . . . . true for any value of x
__
6. The equation is ...
(3x +6) -2x = x +3
x +6 = x +3 . . . . . not true for any number
_____
<em>Additional comment</em>
The work shown is correct. You find x by writing and solving an appropriate equation. To solve the equation, undo the steps that are done to x. Undo addition by adding the opposite. Undo multiplication by diving by the multiplier. It often helps to collect terms before you start the "undo" steps.
For x on both sides of the equation (as in problems 4–6), you can subtract the term with the lowest coefficient. (Of course, you must subtract it from both sides of the equation.) In problem 5, if you do that, you get 1=1, which is true for any value of x. In problem 6, if you do that, you get 6=3, a false equation, indicating no value of x will make it true.