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4 years ago

X-y+z for x= 1,y=6,z=1

Mathematics

1 answer:

adell [148]4 years ago

6 0

Answer:

I'm glad you asked!

Step-by-step explanation:

x = $1.$ So the starting number is 1.

$y=6$ .So it is $1-6$ .

$z = 1$ .So it is now $1 - 6+1$ .

$1-6+1 = -4$ .Your final answer is $-4.$

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Help me determine if relations 1, 2, 3, and 4 are functions or not please!

Reil [10]

They are all not functions

5 0

3 years ago

Suppose the population of a certain city is 5358 thousand. It is expected to decrease to 4565 thousand in 50 years. Find the per

notsponge [240]

Answer:

The population decreases at the rate of 0.32% a year.

Step-by-step explanation:

The population of this certain city can be modeled by this following differential equation.

$\frac{dP}{dt} = Pr$

where r is the growth rate(r>0 means that the population increases, r < 0 it decreases).

We can solve this by the variable separation method. We have that:

$\frac{dP}{P} = rdt$

Integrating both sides, we have

$ln{P} = rt + P(0)$

where P(0) is the initial population.

To find P in function of t, we apply the exponential to both sides.

$e^{ln{P}} = e^{rt + P(0)}$

$P(t) = P(0)e^{rt}$

The initial population of the city 5,358,000. So P(0) = 5,358,000.

It decreases to 4,565,000 in 50 years. So P(50) = 4,565,000.

Applying to the bold equation:

$5,358,000 = 4,565,000e^{50r}$

$e^{50r} = 1.174$

To find the growth rate, we apply ln to both sides.

$ln{e^{50r}} = ln{1.174}$

$50r = 0.16$

r = $\frac{0.16}{50} = 0.0032 = 0.32%$

The population decreases at the rate of 0.32% a year.

5 0

3 years ago

Option 1: $30 an hour plus a non-refundable deposit of $50. Option 2: $20 an hour plus a non-refundable deposit of $75. Kyle is

Margaret [11]

Option 1 equation- 30x+50
Option 2 equation- 20x+75
There is automatically a one time fee of $50 for option 1 and $75 for option 2

1 day:
Option 1- this person pays a total of $80 (because of the one time fee of 50 and 30 per day)
option 2- pays a total of $95 (one time fee of 75 and then an additional 20 per day)

Day 2:
Option 1- $80+$30(daily fee)=$110
Option 2- $95+$20(daily fee)=$115

Day 3:
Option 1- $110+$30=$140
Option 2-$115+$20=$135

They will have to stay a minimum of 3 days for option 2 to be a better deal

7 0

4 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

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

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

6 0

4 years ago

Sonny reads 4 books in 8.5 hours, at this rate how many books does he read in one hour?

Anon25 [30]

You need to figure out how many hours does he need to read one book.

8.5 / 4 = 2.125

Hope this helps! (:

7 0

3 years ago