Triangle town wants to build a school that is equidistant from its three cities L, M, and N. Which construction correctly finds
the best location of the school?
Option choices are...
A. Construction Y because point E is the incenter of ΔLMN
B. Construction Y because point E is the circumcenter of ΔLMN
C. Construction X because point C is the incenter of ΔLMN
D. Construction X because point C is the circumcenter of ΔLMN
Option choices are...
A. Construction Y because point E is the incenter of ΔLMN
B. Construction Y because point E is the circumcenter of ΔLMN
C. Construction X because point C is the incenter of ΔLMN
D. Construction X because point C is the circumcenter of ΔLMN
1 answer:
You might be interested in
Answer:
General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
Derivative Property [Addition/Subtraction]:
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
Integration Rule [Reverse Power Rule]:
Integration Property [Multiplied Constant]:
Integration Methods: U-Substitution and U-Solve
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify given.</em>
<em />
<u>Step 2: Integrate Pt. 1</u>
<em>Identify variables for u-substitution/u-solve</em>.
- Set <em>u</em>:
- [<em>u</em>] Differentiate [Derivative Rules and Properties]:
- [<em>du</em>] Rewrite [U-Solve]:
<u>Step 3: Integrate Pt. 2</u>
- [Integral] Apply U-Solve:
- [Integrand] Simplify:
- [Integral] Rewrite [Integration Property - Multiplied Constant]:
- [Integral] Apply Integration Rule [Reverse Power Rule]:
- [<em>u</em>] Back-substitute:
∴ we have used u-solve (u-substitution) to <em>find</em> the indefinite integral.
---
Learn more about integration: brainly.com/question/27746495
Learn more about Calculus: brainly.com/question/27746485
---
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
Step-by-step explanation:
To prove divisibility, we need to factor the divident such that one of its factors matches the divisor.
(I use the notation x|y to denote that x divides y)
(A)
(B)
In this case, it is easier to also factor the divisor to primes:
Both of these factor must be matched in the dividend in order to prove divisibility, and that indeed turns out to be true: