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2 years ago

Some helppppp!!!!! I’m having trouble

Mathematics

2 answers:

Leno4ka [110]2 years ago

8 0

Oh I was finna answer but he did already

USPshnik [31]2 years ago

3 0

Answer:

1.4*40=56cm, DVD’s he has

1.4*3=4.2cm

56+4.2=60.2cm, whoops too much. So instead 1.4*2= 2.8cm, 56+2.8=58.8cm Is Your Answer.

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SOMEONE HELP ME WITH THIS

Alexandra [31]

Answer:

X=7 and y=6

Step-by-step explanation:

Y=-2x + 20.......eq(i)

6x-5y=12.....eq(ii)

Now, putting y= - 2x +20 in eq (ii)

6x - 5(-2x + 20)=12

6x + 10x - 100 = 12

16x=12+100

X=112 /16

X=7

Then, putting x=7 in eq (i)

Y=-2*7 +20

=-14 +20

8 0

2 years ago

Hi does anyone know the answer

damaskus [11]

Answer:

3.7

Step-by-step explanation:

3 0

3 years ago

Svetlana's hair is 4 centimeters long. Her hair grows 1.5 centimeters per month. Svetlana wants her hair to grow so that it is a

kap26 [50]

It would take her 2.6 months

7 0

3 years ago

Given the differential Equation y'+2y=2e^x ;solve this equation using the integration factor; solve for y to get the general sol

SCORPION-xisa [38]

$y'+2y=2e^x\Longrightarrow y'=2e^x-2y$

If $f'(x)=g(x)$ then $y=\int{g(x)dx}$

So we extract,

$y=\int{2e^x-2x}dx$

Which becomes,

$y=2e^x-x^2+C$

Hope this helps.

r3t40

5 0

3 years ago

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