We
know that
<span>
1) The degree of a polynomial is the highest degree of its terms</span>
<span>2) The leading term in a polynomial is the highest degree term</span>
<span>3) The leading coefficient of a polynomial is the coefficient of the leading term</span>
Therefore
The coefficient
of first term of a polynomial written in standard (descending order) form is the
coefficient of the leading term, thus is called the leading coefficient.
the answer is
The leading coefficient<span>
</span>
Answer:
1
Step-by-step explanation:
<em>look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>
<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
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