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3 years ago

Hellppp asapppp tysmm

Mathematics

2 answers:

Vsevolod [243]3 years ago

4 0

Answer:

C - 3/5

Step-by-step explanation:

add all the numbers up (30, 40, 15, 15) which = 100

subtract 40 from 100 giving you 60

make it a fraction giving you 60/100

simplify your fraction reducing to 3/5

Fiesta28 [93]3 years ago

3 0

Answer:

3/5

Step-by-step explanation:

30 + 40 + 15 + 15 = 100

40/100 will be green

100/100 - 40/100 = 60/100

60/100 will NOT be green

60 ÷ 5 = 12

100 ÷ 5 = 20

12 ÷ 4 = 3

20 ÷ 4 = 5

3/5

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3 years ago

Find each unit rate and determine which is lower.

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3 years ago

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A jet plane can carry up to 200,000 liters of fuel. It used 130,000 liters of fuel during a flight. What percentage of the fuel

Leto [7]

Answer:

Step-by-step explanation:

65%

Step-by-step explanation:

Given: Full capacity= 200000 litres

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Now, finding percentage of fuel used on the flight.

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2 years ago

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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,

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

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

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3 years ago

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Whitepunk [10]

Answer:

Step-by-step explanation:

N = 3rt^4 - 5rz

now factor r out

N = r(3t^4-5z)

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subtract (3t^4)

N/r - 3t^4 = -5z

now divide by -5 to isolate the variable (z)

$\frac{(n/r)-3t^4}{-5}$ = z

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