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2 years ago

What is the area of rectangle PQRS with vertices at P(0,2), Q(0,10), R(4,10), and S(4,2)?

Mathematics

1 answer:

nadezda [96]2 years ago

6 0

It’s 4,2 the and the other on PQRS

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A small rocket is fired from a launchpad 15 m above the ground with an initial velocity left angle 400,450,600 right angle m/s.

vivado [14]

Answer:

v=<400, 450+3t, 600-9.81t> m/s

r=<400t, 450t+3 $t^{2}$ /2, 15+600t-9.81 $t^{2}$ /2> m/s

b) attach file

c) t=122.324s

d) z=18369.9854 m

Step-by-step explanation:

The equation for velocity of each component is v= $v_{0}$ +at, in this case the x-axis component has acceleration 0 $\frac{m}{s^{2} }$ , the y-axis component has acceleration 3 $\frac{m}{s^{2} }$ and the z-axis component has -g $\frac{m}{s^{2} }$ , where g=9.81 $\frac{m}{s^{2} }$ .

The equation for position of each component is x= $x_{0}$ + $v_{0}$ t+a $t^{2}$ /2,

with < $x_{0}$ , $y_{0}$ , $z_{0}$ >=<0,0,15>.

c) This time is t when z=0, with positive sign.

d)Solving the equation $V_f^{2}$ = $V_o^2$ + 2*a*(z- $z_{0}$ ) for z, with $V_f$ =0.

3 0

3 years ago

Martin is paid an hourly rate of d dollars per hour for the first 40 hours he works per week. He is paid one-and-a-half times hi

Nat2105 [25]

Answer:

40d + 10(1.5d) = 770 can be used to determine Martin's hourly pay.

Step-by-step explanation:

Given that:

Hourly rate = d

Earning to first 40 hours = 40d

Earning of more than 40 hours = 1.5d

Amount paid per week = $770

Hours worked = 50 hours

first 40 hours + 10 hours = total earned

40d + 10(1.5d) = 770

Hence,

40d + 10(1.5d) = 770 can be used to determine Martin's hourly pay.

5 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$