Ask question

Ask question

All categories

1 year ago

7

Use the fact that d = rt to write the system of equations that represents the scenario. Let x be the speed of Aro's paddling and

let y be the speed of the river. Upstream: 5.2 = ( ) Downstream: 5.2 = ( )

1 answer:

AnnZ [28]1 year ago

7 0

Answer:

Answer: upstream: 5.2= (x-y) 5

Downstream: 5.2= (x+y) 2.5

Step-by-step explanation:

You might be interested in

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

1 year ago

Evaluate the numerical expression. −8.1 – 3.5 ÷ (−0.5)

beks73 [17]

The answer is -1.1
Hope you get it right

5 0

1 year ago

Loretta deposits $350 every quarter into a savings account that earns 4.5% interest compounded quarterly. What is the balance af

Ann [662]

I am thinking the answer is C

8 0

1 year ago

Find: (4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y)

const2013 [10]

Answer:

The coefficients are

11,-4 and 0

Step-by-step explanation:

We are given two algebraic expressions. We are subtracting the second from the first.

First one is

$4x^2y^3+2xy^2-2y$

Second expression is

$-7x^2y^3+6xy^2-2y$

The given expression

= $4x^2y^3+2xy^2-2y-(-7x^2y^3+6xy^2-2y)\\=4x^2y^3+2xy^2-2y+7x^2y^3-6xy^2+2y\\=11x^2y^3-4xy^2+0y$

The coefficients are

11,-4 and 0

7 0

1 year ago

Read 2 more answers

Over the course of a year,a tractor-trailer commutes 160,000 miles across America.Assuming a trucker changes his tires every 40,

Nikolay [14]

160,000 - 40,000= 120,000/4 = 3 sets of new tires in the first year and 4 every consecutive year.

3 0

1 year ago

Read 2 more answers