Ask question

Ask question

All categories

3 years ago

6

If two men can finish a piece of work in 30 days, then how many men will be required to finish the same work in 10 days?

1 answer:

Bumek [7]3 years ago

4 0

If two men can finish a piece of work in 30 days, then how many men will be required to finish the same work in 10 days?

it would take a half of man to finish the same piece of work in 10 days.

You might be interested in

Calculate the real GDP of the United States for the year 2017. Use 2012 as the base period and

lakkis [162]

2011 because it has the higher gpa of all of them

3 0

3 years ago

Read 2 more answers

4 3/4 divided by 1 1/6

ozzi

Answer:

$2\frac{13}{22}$

Step-by-step explanation:

$=\frac{19}{4}\div \frac{11}{6}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{19}{4}\times \frac{6}{11}$

$=\frac{19}{2}\times \frac{3}{11}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{19\times \:3}{2\times \:11}$

$\mathrm{Multiply\:the\:numbers:}\:19\times \:3=57$

$=\frac{57}{2\times \:11}$

$\mathrm{Multiply\:the\:numbers:}\:2\times \:11=22$

$=\frac{57}{22}$

$=2\frac{13}{22}$

5 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

−2/3a+1/8−1/6a−3/4 PLEASE HELP

prohojiy [21]

Answer:

Evaluate:

- $\frac{5a}{6}$ - $\frac{5}{8}$

Factor:

$\frac{5(-4a-3}{24}$

Step-by-step explanation:

6 0

3 years ago

Two mechanics worked on a car. The first mechanic worked for 20 hours, and the second mechanic worked for 15 hours. Together the

klasskru [66]

The rate charged per hour by each mechanic was: x = 75 $ / hr and y = 115 $ / hr.

<h3>What is a system of equations?</h3>

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

Given;

The first mechanic worked for 20 hours, and the second mechanic worked for 15 hours.

Together they charged a total of $3225.

For this case we have the following variables:

x be the amount of $ / hr that the mechanic obtains 1.

y be the amount of $ / hr obtained by mechanic 2.

An equation to express this would be:

x + y = 190

20x + 15y = 3225

Solving the system of equations we have:

20x + 15(190 -x) = 3225

20x + 2850 - 15x = 3225

5x = 375

x = 75

simililary

y = 190 - x

y = 115

Hence, the rate charged per hour by each mechanic was:

x = 75 $ / hr

y = 115 $ / hr

Learn more about equations here;

brainly.com/question/10413253

#SPJ1

6 0

2 years ago