Answer:
1440 ft squared
Step-by-step explanation:
v = lwh (length times with time height)
Hope this helps Buddy!
- Courtney
Answer: <em>Hey, It's me, Lily!!</em>
Step-by-step explanation:
1) x = −5/2
2) x = 4
3) x = 5
4) x = 6
5) x = 7
<u>Hope This Helps!! (^ '--')</u>
Answer:
Area of Trapezoid is 39 unit²
Step-by-step explanation:
Given as :
For A Trapezoid
The measure of base side 1 =
= 10 unit
The measure of base side 2 =
= 16 unit
The height of the Trapezoid = h = 3 unit
Let The Area of Trapezoid = A square unit
<u>Now, From Formula</u>
Area of Trapezoid =
× (sum of opposite base) × height
I.e A =
× (
+
) × h
Or, A =
× (10 unit + 16 unit) × 3 unit
Or, A =
× (26 unit) × 3 unit
Or, A =
× 78 unit²
Or, A =
unit²
I.e A = 39 unit²
So, The Area of Trapezoid = A = 39 unit²
Hence, The Area of Trapezoid is 39 unit² . Answer
Answer:
B) 4√2
General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Parametric Differentiation
Integration
- Integrals
- Definite Integrals
- Integration Constant C
Arc Length Formula [Parametric]: ![\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20AL%20%3D%20%5Cint%5Climits%5Eb_a%20%7B%5Csqrt%7B%5Bx%27%28t%29%5D%5E2%20%2B%20%5By%28t%29%5D%5E2%7D%7D%20%5C%2C%20dx)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>

Interval [0, π]
<u>Step 2: Find Arc Length</u>
- [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:

- Substitute in variables [Arc Length Formula - Parametric]:
![\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20AL%20%3D%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%20%7B%5Csqrt%7B%5B1%20%2B%20sin%28t%29%5D%5E2%20%2B%20%5B-cos%28t%29%5D%5E2%7D%7D%20%5C%2C%20dx)
- [Integrand] Simplify:
![\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20AL%20%3D%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%20%7B%5Csqrt%7B2%5Bsin%28x%29%20%2B%201%5D%7D%20%5C%2C%20dx)
- [Integral] Evaluate:
![\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20AL%20%3D%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%20%7B%5Csqrt%7B2%5Bsin%28x%29%20%2B%201%5D%7D%20%5C%2C%20dx%20%3D%204%5Csqrt%7B2%7D)
Topic: AP Calculus BC (Calculus I + II)
Unit: Parametric Integration
Book: College Calculus 10e