(3+5.50)5
8.50*5=42.5
Derek spends $42.50 each school week.
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
Answer:
Quantity of heat needed (Q) = 722.753 × 10³
Step-by-step explanation:
According to question,
Mass of water (m) = 40 kg
Change in temperature ( ΔT) = 18°c
specific heat capacity of water = 4200 j kg^-1 k^-1
The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .
So, Heat (Q) = m×s×ΔT
Or, Q = 40 kg × 4200 × 18
or, Q = 3024 × 10³ joule
Hence, Quantity of heat needed (Q) = 3024 × 10³ joule
In calories 4.184 joule = 1 calorie
So, 3024 × 10³ joule = 722.753 × 10³
Answer:
Step-by-step explanation:
The second one I believe.