At some point, the exponential function always exceeds a polynomial function.
9514 1404 393
Answer:
2√30 ∠-120°
Step-by-step explanation:
The modulus is ...
√((-√30)² +(-3√10)²) = √(30 +90) = √120 = 2√30
The argument is ...
arctan(-3√10/-√30) = arctan(√3) = -120° . . . . a 3rd-quadrant angle
The polar form of the number can be written as ...
(2√30)∠-120°
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<em>Additional comments</em>
Any of a number of other formats can be used, including ...
(2√30)cis(-120°)
(2√30; -120°)
(2√30; -2π/3)
2√30·e^(i4π/3)
Of course, the angle -120° (-2π/3 radians) is the same as 240° (4π/3 radians).
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At least one app I use differentiates between (x, y) and (r; θ) by the use of a semicolon to separate the modulus and argument of polar form coordinates. I find that useful, as a pair of numbers (10.95, 4.19) by itself does not convey the fact that it represents polar coordinates. As you may have guessed, my personal preference is for the notation 10.95∠4.19. (The lack of a ° symbol indicates the angle is in radians.)
Answer:
y = -2x
Step-by-step explanation:
For direct variation: y = kx, where k is the constant of proportionality.
Given that Y equals -12 when x equals 6, we have -12 = k(6). Solve this for k by dividing both sides by 6:
k = -12/6 => k = -2
Then the "direct variation equation" here is y = -2x.
Answer:
A) mPQ = 71º
B) mSR = 161º
C) mQRT = 199ª
D) mPSR = 270º
E) mPS = 109º
Step-by-step explanation:
We know that mST is 19º and QR is also 19º because they're opposite angles. We also know that angle PUR is 90º.
If we subtract 19º from 90º we get 71º for mPQ.
We also know that TR is a 180º angle. Using this if we take 19º from 180º we're left with 161º for mSR.
Using the 180º from angle TR, as well as the 19º from mQR we know that mQRT has to be 199º.
A circle is 360º and a right angle is 90º. That means that mPSR is 270º.
Knowing that TR is 180º and that PR is 90º, PTº must be supplementary making it also 90º. Adding the 19º from ST to the 90º from TP we know that PS is 109º
