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11111nata11111 [884]
3 years ago
12

For what value of x must ABCD be a​ parallelogram?

Mathematics
2 answers:
bija089 [108]3 years ago
8 0

5x = 6x-7

subtract 6x from each side

-x=-7

x=7


aksik [14]3 years ago
8 0

Answer: The value of x must be 7 units.

Step-by-step explanation:

Since we have given that

ABCD is a parallelogram.

Suppose that AC and BD intersect at o.

So, AO = OC.

And we have given that

AO = 5x

OC = 6x-7

According to question,

5x=6x-7\\\\5x-6x=-7\\\\-x=-7\\\\x=7

Hence, the value of x must be 7 units.

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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)
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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
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  • 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.
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  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
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