If 52 men can do a piece of work in 35 days in how many days will 28 men to it?

Mathematics

2 answers:

dlinn [17]3 years ago

8 0

28 mando the peace of work in 41 days

by divida 52 and 35 and multiiply with28

aniked [119]3 years ago

7 0

One of them have the virus they all get infected.

So they make none unless the manager tries the factory is closed down now

You might be interested in

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

The volume

Sedaia [141]

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&& y={{ k }}x
\end{array}\\ \quad \\
$

and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}
$

now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

so $\bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
P=10
\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T$