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kenny6666 [7]
2 years ago
11

What is 1/4 written as a percent

Mathematics
2 answers:
olga nikolaevna [1]2 years ago
4 0
Tue answer is 1/4 written as a percent is 25% hope this helps you
frosja888 [35]2 years ago
3 0
1/4 written as a percent is 25%. To get this answer, divide 1 into 4. 1/4=0.25. Now, multiply 0.25 by 100 to get 25. :) 
Hope this helps! 
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Name an angle bisector shown in the figure.<br><br><br> DE<br><br> AB<br><br> GB<br><br> CF
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Answer:

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Step-by-step explanation:

we know that

An <u><em>angle bisector</em></u> is a line that divide an angle into two equal angles.

In this problem

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What is the solution to the compound inequality in interval notation?
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3 years ago
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What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)
zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. 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

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.

Interval [0, π]

<u>Step 2: Find Arc Length</u>

  1. [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:         \displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.
  2. Substitute in variables [Arc Length Formula - Parametric]:                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx
  3. [Integrand] Simplify:                                                                                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx
  4. [Integral] Evaluate:                                                                                         \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0
2 years ago
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