Answer:
u have some very good handwriting
4) (a) For these problems, you should take time to familiarize yourself with common fractions that appear on the unit circle.
does not appear in the unit circle unless you take the quotient 1/2 divided by sqrt(3)/2 which gives you 1/sqrt(3) which is the same as sqrt(3)/3. So our numerator is 1/2 and our denominator is sqrt(3)/2.
And remember tangent is just sin/cos. So what degree has sinx as 1/2 and and cosx as sqrt(3)/2? Well, 30 degrees does, but 30 degrees is not within the range we are given. That means they are looking for a sinx that gives us -1/2 and a cosx that gives us -sqrt(3)/2 and that is 210 degrees.
And 210 degrees in radians is 7pi/6.
I hoped that made sense.
(b) This is a lot easier. What angle gives us a cos x of -sqrt(3)/2? According to the unit circle, 150 degrees and 210 degrees does. They usually want these in radians, so the answer is 5pi/6 and 7pi/6, respectively.
5) What quadrant is radian measure 5 in?
Well 2pi or roughly 6.28 is a full circle. And 5 is slightly less than 6.28, so it is probably in quadrant IV.
But to be sure let's change 5 radian to degrees:
5 * 180/pi = 900/pi = 286.48 degrees
286.48 degrees is definitely in Q4, so we are correct.
Answer:
The volume of tank B is 33.7 cubic centimeters greater than the volume of tank A.
Step-by-step explanation:
Volume of tank A 530(18.4)= 9,752
Volume of tank B (26.2)(16.6)(22.5) =9,785.7
Volume of tank B - Volume of tank A 9,785.7 - 9,752 = 33.7
Answer:
If I calculated correctly, the tangent line is horizontal where x ≈ -5.3 + 9.3i, and -5.3 - 9.3i
I'm somewhat concerned at having gotten complex numbers, and strongly recommend going through the steps to see if I missed anything. I checked it myself and don't see any errors.
Step-by-step explanation:
You can do this by taking the derivative of the function and solving for zero:
f(x) = 2x³ + 32x² + 220x + 11
f'(x) = 6x² + 64x + 220
f'(x) = 2(3x² + 32x + 110)
We can't factor that further, so let's do it the long way, starting by letting f'(x) equal zero:
0 = 2(3x² + 32x + 110)
0 = 3x² + 32x + 110
0 = 9x² + 96x + 990
0 = 9x² + 96x + 256 + 734
0 = (3x + 16)² + 734
(3x + 16)² = -734
3x + 16 = ± i√734
3x = -16 ± i√734
x = (-16 ± i√734) / 3
x ≈ (-16 + 27.9i) / 3, and (16 - 27.9i) / 3
x ≈ -5.3 + 9.3i, and -5.3 - 9.3i
I'm always wary when I end up with complex numbers. I'd suggest double checking everything here, but I'm fairly certain I did everything correctly.