Answer:
0 as an exponent would look something like 6^0 and anything to the power of 0 is always 1
Answer:

General Formulas and Concepts:
<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)
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
Integration Rule [Fundamental Theorem of Calculus 1]: 
Integration Property [Multiplied Constant]: 
U-Substitution
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Integrate Pt. 1</u>
<em>Identify variables for u-substitution.</em>
- Set <em>u</em>:

- [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:

- [Bounds] Switch:

<u>Step 3: Integrate Pt. 2</u>
- [Integral] Rewrite [Integration Property - Multiplied Constant]:

- [Integral] U-Substitution:

- [Integral] Exponential Integration:

- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:

- Simplify:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
x = 28 cm
Step-by-step explanation:
Given:
Area of link shaded regions = 84 cm²
Required:
The value of x (diameter of the semicircle/length of the rectangle)
Solution:
Diameter of the semicircle = 2r = x
Length of rectangle (L) = 2r = x
Radius of semicircle (r) = ½x
Width of rectangle (W) = radius of semicircle = ½x
Use 3.14 as π
Area of the link shaded regions = area of rectangle - area of semicircle
Thus:
Area of the link shaded regions = (L*W) - (½*πr²)
Plug in the values
84 = (x*½x) - (½*3.14*(½x)²)
84 = x²/2 - (1.57*x²/4)
84 = x²(½ - 1.57/4)
84 = x²(0.5 - 0.3925)
84 = x²(0.1075)
Divide both sides by 0.1075
84/0.1075 = x²
781.4 = x²
√781.4 = x
27.9535329 = x
x = 28 cm
Answer:
Image result for The length of a rectangle is 3 more than five times its width. If the perimeter of the rectangle is 30 meters, find the dimensions of the rectangle.
To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides. This number divided by 2 is the width.