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
655,822
Please give brainliest
a) Since the corresponding y-value is -0.6, hence the point (-0.8, -0.6) is a solution to the system of equations
b) since the corresponding x-value is not 1/3, hence the point (1/3, 2) is not a solution to the system of equation
In order to show if the given point corresponds to the given function, we will have to substitute the value of x into the function to see if we will have its corresponding y-value
For the point (-0.8, -0.6), substitute x = -0.8 into both functions as shown:
f(x) = 2x + 1
f(-0.8) = 2(-0.8) + 1
f(-0.8) = -1.6 + 1
f(-0.8) = -0.6
Simiarly;
y = -3(-0.8)- 3
y = 2.4 - 3
y = -0.6
Since the corresponding y-value is -0.6, hence the point (-0.8, -0.6) is a solution to the system of equations
For the point (1/3, 2), substitute x = 1/3 into both functions as shown:
x = (y+2)/2
x = (2+2)/2
x = 4/2
x = 2
Simiarly;
x + 2 = 3
x = 3-2
x = 1
Since the corresponding x-value is not 1/3, hence the point (1/3, 2) is not a solution to the system of equations
Learn more on systems of equation here: brainly.com/question/847634