Answer:
84 ![cm^{2}](https://tex.z-dn.net/?f=cm%5E%7B2%7D)
Step-by-step explanation:
find the area of each shape and add them together
Answer:
See explanation.
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Functions
- Exponential Property [Rewrite]:
![\displaystyle b^{-m} = \frac{1}{b^m}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20b%5E%7B-m%7D%20%3D%20%5Cfrac%7B1%7D%7Bb%5Em%7D)
- Exponential Property [Root Rewrite]:
![\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: ![\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bcf%28x%29%5D%20%3D%20c%20%5Ccdot%20f%27%28x%29)
Derivative Property [Addition/Subtraction]: ![\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%20%2B%20g%28x%29%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5D%20%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%29%5D)
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: ![\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
Step-by-step explanation:
We are given the following and are trying to find the second derivative at <em>x</em> = 2:
![\displaystyle f(2) = 2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%282%29%20%3D%202)
![\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%206%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D)
We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:
![\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%206%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B1%7D%7B2%7D%7D)
When we differentiate this, we must follow the Chain Rule: ![\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%206%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5CBig%5D%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%20%28x%5E2%20%2B%203y%5E2%29%20%5CBig%5D)
Use the Basic Power Rule:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%282x%20%2B%206yy%27%29)
We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%206y%286%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%29%20%5Cbig%5D)
Simplifying it, we have:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D)
We can rewrite the 2nd derivative using exponential rules:
![\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%7D)
To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:
![\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%5Cbigg%7C%20%5Climits_%7Bx%20%3D%202%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202%282%29%20%2B%2036%282%29%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%7D)
When we evaluate this using order of operations, we should obtain our answer:
![\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%5Cbigg%7C%20%5Climits_%7Bx%20%3D%202%7D%20%3D%20219)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
They are dependent events because if cara spins it and it lands on her she gets to play with the puppy, But if she misses it and her brother spins and it lands on him he gets to play with the puppy.
Short answer: the event is dependent because the outcome of the first spin effects the brother outcome
Answer:
Step-by-step explanation:
Option 1:
4(m + 12) = 4m + 48 (not what we want)
Option 2:
6m - 2m + 12 = 4m + 12 (one of the right answers)
Option 3:
4(m + 3) = 4m + 12 (one of the right answers)
Option 4:
3m - m + 12 = 2m + 12 (not what we want)
Therefore the correct answers are Option 2 and Option 3.
Answer:
The trapezoid.
Step-by-step explanation:
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