5/12, 2/3, 11/12, 7/6, 17/12 find the common difference between the terms

Softa [21]

The answer to this question is B!

8 0

3 years ago

Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisti

timofeeve [1]

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

$\bf \int_{C}F$

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write

C = $\bf C_1+C_2$

where

$\bf C_1$ = line segment from (1, 5) to (0, 0)

$\bf C_2$ = line segment from (0, 0) to (0, 9)

so,

$\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F$

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1

Hence $\bf C_1$ can be parameterized as

$\bf r_1(t) = (1-t, 5-5t)$ for 0 ≤ t ≤ 1

and $\bf C_2$ can be parameterized as

$\bf r_2(t) = (0, 9t)$ for 0 ≤ t ≤ 1

The derivatives are

$\bf r_1'(t) = (-1, -5)$

$\bf r_2'(t) = (0, 9)$

and

$\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0$

$\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0$

In consequence,

$\bf \int_{C}F=0$

6 0

4 years ago

Do you like quarintine

agasfer [191]

Answer:

no, I don't like quarantine.

x should be 115 because it is parallel to the one above. each line should equal 180. So 180-115= 65

y=65

4 0

4 years ago

Read 2 more answers

Need help finding the missing angle

lys-0071 [83]

Answer:

Step-by-step explanation:

Sinθ = opp/hyp

Sinθ = 12/13

θ = ArcSin (12/13)

θ =67.38

Let x = adj

Tan (67.38) = 15/x

2.4= 15/x

x2.4= 15

x=6.25

Therefore the missing value is 6.25

6 0

2 years ago

Help with num 1 please.

KengaRu [80]

Answer:

(i) $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii) $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii) $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Exponential Differentiation

Logarithmic Differentiation

Step-by-step explanation:

(i)

Step 1: Define

Identify

$\displaystyle y = (3x^2 - x)ln(2x + 1)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (3x^2 - x)'ln(2x + 1) + (3x^2 - x)[ln(2x + 1)]'$
Basic Power Rule/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{1}{2x + 1}(2x + 1)'$
Basic Power Rule: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{2}{2x + 1}$
Simplify [Factor]: $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii)

Step 1: Define

Identify

$\displaystyle y = \frac{x^2 + 2}{lnx}$

Step 2: Differentiate

Quotient Rule: $\displaystyle y' = \frac{(x^2 + 2)'lnx - (x^2 + 2)(lnx)'}{(lnx)^2}$
Basic Power Rule/Logarithmic Differentiation: $\displaystyle y' = \frac{2xlnx - (x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Rewrite: $\displaystyle y' = \frac{2xlnx}{(lnx)^2} - \frac{(x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Simplify: $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii)

Step 1: Define

Identify

$\displaystyle y = e^xln(2x)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'$
Exponential Differentiation/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'$
Basic Power Rule: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2$
Simplify: $\displaystyle y' = e^xln(2x) + \frac{e^x}{x}$
Rewrite: $\displaystyle y' = \frac{xe^xln(2x) + e^x}{x}$
Factor: $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

6 0

3 years ago