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MrMuchimi
2 years ago
10

Irracionais menores que √5

Mathematics
1 answer:
just olya [345]2 years ago
6 0

Answer:

2.23 this the answer use the calculator

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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
3 years ago
Identify the zeros of the function f(x) = 4x2 − 8x − 1 using the Quadratic Formula. HELP ASAP!!
forsale [732]

1+\dfrac{\sqrt{5}}{2},1-\dfrac{\sqrt{5}}{2}

Step-by-step explanation:

The given equation is 4x^{2}-8x-1

Let a be the coefficient of x^{2}

Let b be the coefficient of x

Let c be the constant.

Then the roots α,β for the equation ax^{2}+bx+c are \dfrac{-b+\sqrt{b^{2}-4ac} }{2a},\dfrac{-b-\sqrt{b^{2}-4ac} }{2a}

So,α=\frac{-b+\sqrt{b^{2}-4ac} }{2a}=\frac{8+\sqrt{64+16} }{8}=\frac{8+4\sqrt{5}}{8}=1+\frac{\sqrt{5}}{2}

β=\frac{-b-\sqrt{b^{2}-4ac} }{2a}=\frac{8-\sqrt{64+16} }{8}=\frac{8-4\sqrt{5}}{8}=1-\frac{\sqrt{5}}{2}.

So the roots are 1+\frac{\sqrt{5}}{2},1+\frac{\sqrt{5}}{2}

5 0
3 years ago
Simplify:<br> -11√21-11√/21
alex41 [277]

Step-by-step explanation:

by using the radical and suds rule

6 0
2 years ago
What is the given point of m= -2, (2,5)
kap26 [50]

Answer: i think it's (-4,-10)=m

Step-by-step explanation:

5 0
3 years ago
3 cm
dolphi86 [110]

Answer:

D.72cm²

Step-by-step explanation:

sa=5(s²)+(s×w)+(L×s)+(s×p)

sa=5(3²)+(3×6)+(2×3)+(3×1)

sa=45cm²+8cm²+6cm²+3cm²

sa=72cm²

3 0
2 years ago
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