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2 years ago

6. The variable n represents the figure number

Mathematics

1 answer:

vlabodo [156]2 years ago

7 0

Brodsky I wondering the same thing but good luck finding the answer

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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:

Calculus

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]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

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

Interval [0, π]

Step 2: Find Arc Length

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

(2y^3+5y^2-y+15) ÷ (y-4)

Sveta_85 [38]

ANSWER: 2y^2 + 13y +51 r(219/y-4)

picture explanation too:

8 0

2 years ago

Use these functions to answer the questions.

Elenna [48]

Part A
We have $f\left(x\right)=\left(x+3\right)^2-1$ . To solve for the x-intercept, we set f(x) equal to 0. That is
$\left(x+3\right)^2-1=0$

$\left(x+3\right)^2=1$

Take the square root of both sides,
$x+3=1$

$x=-2$

The x-intercept is (-2,0).

To solve for the y-intercept, we set x=0. That is
$y=\left(0+3\right)^2-1=3^2-1=9-1=8$

The y-intercept is (0, 8)

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have $g\left(x\right)=-2x^2+8x+3$ .
Factor out -2
$=-2\left(x^2-4x\right)+3$

Complete the square
$=-2\left(x^2-4x+4\right)+3-2\left(4\right)$

Simplify
$g(x)=-2\left(x-2\right)^2-5$

Part C
We have $h\left(t\right)=-16t^2+28t$ .
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo.