Compute the area (in Acres) of a rectangular parcel of land having length of 200.013 USft and width of 147.631 USft:

__________ acres

__________ acres

Elianna Lawrence
2022-07-28
Answered

__________ acres

You can still ask an expert for help

Tamoni5e

Answered 2022-07-29
Author has **14** answers

Solution:

Area of rectangular parcel of land = Length X Width

= 200.013 x 147.631

= 29528.1192 acres

Area of rectangular parcel of land = Length X Width

= 200.013 x 147.631

= 29528.1192 acres

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$max\text{}min[\alpha {x}_{1},\beta {x}_{2},\gamma {x}_{3}]\text{}\text{}\text{s.t.}\text{}{\lambda}_{1}{x}_{1}+{\lambda}_{2}{x}_{2}+{\lambda}_{3}{x}_{3}=c,\phantom{\rule{0ex}{0ex}}\text{}\alpha ,\beta ,\gamma ,{\lambda}_{i},c\text{}\text{are constants}$

Well, that function is not differentiable , so what methods can be applied to solve for for the optimal values of ${x}_{1}$,${x}_{2}$ and ${x}_{3}$? Is knowledge of the ${\lambda}^{\prime}s$ and $c$ necessary, to at least some degree, or does a general approach/solution exist?

Well, that function is not differentiable , so what methods can be applied to solve for for the optimal values of ${x}_{1}$,${x}_{2}$ and ${x}_{3}$? Is knowledge of the ${\lambda}^{\prime}s$ and $c$ necessary, to at least some degree, or does a general approach/solution exist?

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Problem with finding the volume with cylindrical coordinates.

I have the following function:

$Z={x}^{2}+{y}^{2},\text{}Z=x+y$

that i want to solve (to find the volume) with cylindrical coordinates. I am evaluating the integral to get:

$V=\iiint dV=\underset{R}{\iint}\phantom{\rule{-5pt}{0ex}}{\int}_{{x}^{2}+{y}^{2}}^{x+y}\phantom{\rule{thinmathspace}{0ex}}dz\phantom{\rule{thinmathspace}{0ex}}dA=\underset{R}{\iint}x+y-({x}^{2}+{y}^{2})\phantom{\rule{thinmathspace}{0ex}}dA.$

and from here i am trying to get the bounds for r by intersecting the two functions and i get that $r=0\text{}or\text{}r=1$ therefore i tought that $0\le r\le 1$ but it dosen't seems right. But i get the following result:

$2\pi {\int}_{0}^{1}(r-{r}^{2})rdrd\theta =2\pi {\textstyle (}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4}{\textstyle )}{{\textstyle |}}_{0}^{1}=\frac{\pi}{6}$

and this is not the corrent answer, i should get $\frac{\pi}{8}.$. I know this is too specific to a given problem question i apologize for that but can someone tells me where i made a mistake

I have the following function:

$Z={x}^{2}+{y}^{2},\text{}Z=x+y$

that i want to solve (to find the volume) with cylindrical coordinates. I am evaluating the integral to get:

$V=\iiint dV=\underset{R}{\iint}\phantom{\rule{-5pt}{0ex}}{\int}_{{x}^{2}+{y}^{2}}^{x+y}\phantom{\rule{thinmathspace}{0ex}}dz\phantom{\rule{thinmathspace}{0ex}}dA=\underset{R}{\iint}x+y-({x}^{2}+{y}^{2})\phantom{\rule{thinmathspace}{0ex}}dA.$

and from here i am trying to get the bounds for r by intersecting the two functions and i get that $r=0\text{}or\text{}r=1$ therefore i tought that $0\le r\le 1$ but it dosen't seems right. But i get the following result:

$2\pi {\int}_{0}^{1}(r-{r}^{2})rdrd\theta =2\pi {\textstyle (}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4}{\textstyle )}{{\textstyle |}}_{0}^{1}=\frac{\pi}{6}$

and this is not the corrent answer, i should get $\frac{\pi}{8}.$. I know this is too specific to a given problem question i apologize for that but can someone tells me where i made a mistake

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I am stuck on the above practice question: If I say $R=0,1/4,2/4,\dots $… and then use the continuous case equivalent of geometric distribution to find the probability function of R, which is exponential distribution, will I be correct?

Let X have a geometric distribution with $f(x)=p(1-p)x$, $x=0,1,2,\dots $. Find the probability function of R, the remainder when X is divided by 4.

I am stuck on the above practice question: If I say $R=0,1/4,2/4,\dots $… and then use the continuous case equivalent of geometric distribution to find the probability function of R, which is exponential distribution, will I be correct?