Answer:
Point A(9, 3)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Coordinates (x, y)
- Functions
- Function Notation
- Terms/Coefficients
- Anything to the 0th power is 1
- Exponential Rule [Rewrite]:
- Exponential Rule [Root Rewrite]:
<u>Calculus</u>
Derivatives
Derivative Notation
Derivative of a constant is 0
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:
<u>Step 1: Define</u>
<em>Identify</em>
<em />
<em />
<em />
<em />
<em />
<u>Step 2: Differentiate</u>
- [Function] Rewrite [Exponential Rule - Root Rewrite]:
![\displaystyle y = x^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
- Basic Power Rule:
![\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B1%7D%7B2%7Dx%5E%7B%5Cfrac%7B1%7D%7B2%7D%20-%201%7D)
- Simplify:
![\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B1%7D%7B2%7Dx%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D)
- [Derivative] Rewrite [Exponential Rule - Rewrite]:
![\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B1%7D%7B2x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
- [Derivative] Rewrite [Exponential Rule - Root Rewrite]:
![\displaystyle y' = \frac{1}{2\sqrt{x}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D)
<u>Step 3: Solve</u>
<em>Find coordinates of A.</em>
<em />
<em>x-coordinate</em>
- Substitute in <em>y'</em> [Derivative]:
![\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B6%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D)
- [Multiplication Property of Equality] Multiply 2 on both sides:
![\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%7D%7D)
- [Multiplication Property of Equality] Cross-multiply:
![\displaystyle \sqrt{x} = 3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7Bx%7D%20%3D%203)
- [Equality Property] Square both sides:
![\displaystyle x = 9](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%20%3D%209)
<em>y-coordinate</em>
- Substitute in <em>x</em> [Function]:
![\displaystyle y = \sqrt{9}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20%5Csqrt%7B9%7D)
- [√Radical] Evaluate:
![\displaystyle y = 3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%203)
∴ Coordinates of A is (9, 3).
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Derivatives
Book: College Calculus 10e
Let 27 be x% of 60
so,
x/100 *60 = 27
x = 27*100/60 = 270/6 = 45
So,
27 is <em>45%</em> of 60.
Since there are no specific requirements given other than expression written using exponents, I am just going to give some examples.
1. 5x^2 - 3
2. 8^5 - 15x + 7
3. 15x^10 - 8x^7 + 16x^5 + 17^3 - 4^2 +13x - 2
Let's say the side length of square A is x, which means the side length of square B is 2x.
Then, the area of square A can be written as , and the area of square B can be written as .
There's no diagram here with shaded region, so I'll just find the area of square A as a percentage of the area of square B:
= 1/4 = 25%
So, the answer is 25% (note this is the answer to the question: "express the area of square A as a percentage of the area of square B; there is no diagram showing me where the shaded area is, so I cannot answer the original question