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pogonyaev
2 years ago
6

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:

a)

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

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

b) attach file

c) t=122.324s

d) z=18369.9854 m

Step-by-step explanation:

a)

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+at^{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

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

8 0
3 years ago
3/4 x 1/8 in simples form
Zielflug [23.3K]

Answer:

3/32

Step-by-step explanation:

To get this answer, multiply the numerators together and the denominators together

3x1=3

4x8=32

3/32

5 0
3 years ago
What is the correct method to label a line with point U and W on it
Anuta_ua [19.1K]

Answer:

The answer is B.

Step-by-step explanation:

7 0
2 years ago
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