Answer : Slope = -4x/1
Explanation : The line is decreasing, which makes the slope a negative. And if you go down four spaces and to the right one space you have your slope.
Oh okay
Brainliest would really help :))
Answer:
{d,b}={4,3}
Step-by-step explanation:
[1] 11d + 17b = 95
[2] d + b = 7
Graphic Representation of the Equations :
17b + 11d = 95 b + d = 7
Solve by Substitution :
// Solve equation [2] for the variable b
[2] b = -d + 7
// Plug this in for variable b in equation [1]
[1] 11d + 17•(-d +7) = 95
[1] -6d = -24
// Solve equation [1] for the variable d
[1] 6d = 24
[1] d = 4
// By now we know this much :
d = 4
b = -d+7
// Use the d value to solve for b
b = -(4)+7 = 3
Solution :
{d,b} = {4,3}
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