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2 years ago

Parallelogram ABCD is a rectangle. What are the slopes of the sides that make this quadrilateral a rectangle?

Mathematics

1 answer:

snow_tiger [21]2 years ago

6 0

Answer: -3 and 1/3

Step-by-step explanation:

There’s your answer

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levacccp [35]

Answer:

ON = KJ

Step-by-step explanation:

JKL = NOP

We know the angles match

<J = <N

<K = <O

< L = <P

And we know

JK = NO

KL = OP

JL = NP

We are looking for ON = KJ

5 0

2 years ago

Read 2 more answers

How many interior angles does a triangle have

Aleonysh [2.5K]

There are 3 interior angles.

7 0

3 years ago

Read 2 more answers

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Plse help me dawdawdawd

Maksim231197 [3]

12+6x was subtracted from the left but only 12 was subtracted from the right.

6x and 6 are not equal unless x=1

X=-11/15

3 0

2 years ago

Y = x + 3<br>2y = 2x + 6

Marina CMI [18]

Answer:

x=0

y=3

Step-by-step explanation:

2(x+3)=2x+6, substituting the first equation in the second

2x+6=2x+6, rearranging the equations

2x-2x=6-6

0=0

y=0+3

y=3

6 0

3 years ago