Answer:
1. ∠A and ∠B are right angles. Given
2. m∠A = m∠ B All right angles are congruent.
3. ∠BEC≅ ∠AED Vertical angles are congruent
4. ΔCBE ~ ΔDAE AA
Step-by-step explanation:
A proof always begins with the givens.
1. ∠A and ∠B are right angles. -------------->Given
2. m∠A = m∠ B are equal since-----------> All right angles are congruent.
3. ∠BEC≅ ∠AED are also equal since---->Vertical angles are congruent
4. ΔCBE ~ ΔDAE since two angles are equal----------> AA
F(x) = 9x²<span> - 5x + 2
</span>f(-2) = 9(-2)² - 5(-2) + 2
<span>f(-2) = 36 + 10 + 2
</span><span>f(-2) = 48</span>
Answer:

General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: ![\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bf%28x%29g%28x%29%5D%3Df%27%28x%29g%28x%29%20%2B%20g%27%28x%29f%28x%29)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Differentiate</u>
- [Function] Derivative Rule [Product Rule]:
![\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B9x%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Rewrite [Derivative Property - Multiplied Constant]:
![\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%209%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Basic Power Rule:
![\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%2090x%5E9%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Arctrig Derivative:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Answer:
Step-by-step explanation:
It is convenient to memorize the trig functions of the "special angles" of 30°, 45°, 60°, as well as the way the signs of trig functions change in the different quadrants. Realizing that the (x, y) coordinates on the unit circle correspond to (cos(θ), sin(θ)) can make it somewhat easier.
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<h3>20.</h3>
You have memorized that cos(x) = (√3)/2 is true for x = 30°. That is the reference angle for the 2nd-quadrant angle 180° -30° = 150°, and for the 3rd-quadrant angle 180° +30° = 210°.
Cos(x) is negative in the 2nd and 3rd quadrants, so the angles you're looking for are
150° and 210°
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<h3>Bonus</h3>
You have memorized that sin(π/4) = √2/2, and that cos(3π/4) = -√2/2. The sum of these values is ...
√2/2 + (-√2/2) = 0
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<em>Additional comments</em>
Your calculator can help you with both of these problems.
The coordinates given on the attached unit circle chart are (cos(θ), sin(θ)).