Use pythagorean theorem:
sin^2 + cos^2 = 1
(-1/3)^2 + cos^2 = 1
1/9 + cos^2 = 1
cos^2 = 8/9
cos = +- sqrt(8)/sqrt(9) = +- 2sqrt(2)/3
Determine whether cos is positive or negative by looking at which quadrant the angle is in.
pi < theta < 3pi/2 ---> this is 3rd quadrant where x or cos is negative
Therefore cos(theta) = -2sqrt(2)/3
Answer:
- f'(1) exists: f'(1) = -2
- f'(0) DNE
Step-by-step explanation:
<h3>a)</h3>
The function is continuous for x > 0 . The derivative is defined on that interval and is equal to ...
f'(x) = -2x . . . . . for x > 0
Then at x = 1, the derivative is ...
f'(1) = -2(1)
f'(1) = -2
__
<h3>b)</h3>
The function has a jump discontinuity at x=0, so the derivative does not exist at that point. A condition for the existence of the derivative is that the function is continuous at the point of interest.
Answer:
x = 3
Step-by-step explanation:
First set up the problem like this:
14x-17 = 8x + 1
Now we subtract 14x from each side. This leaves us with:
-17 = 6x+1
Now we need subtract 1 from each side, leaving us with:
-18 = -6x
And now we divide each side by 6x, leaving us with.
3 = x.
And if you insert three into both of the equations, it will result in 25.
(Hope this helped!)
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