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2 years ago

11

Bipasha can eat 24 rasgullas in 12 minutes. Her friend Sujata takes DOUBLE the time to eat the same number.One day they buy 24 r

asgullas and start eating them. If they eat at their normal speeds till the rasgullas are over, how many rasgullas would Sujata have eaten?

1 answer:

Crazy boy [7]2 years ago

5 0

Answer:

12

Step-by-step explanation:

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Determine whether each of the following functions is a solution of Laplace's equation uxx + uyy = 0. (Select all that apply.) u

Naddika [18.5K]

Answer with Step-by-step explanation:

We are given that Laplace's equation

$u_{xx}+u_{yy}=0$

We have to determine given function is solution of given laplace's equation.

If a function is solution of given Laplace's equation then it satisfy the solution.

1. $u=e^{-x}cosy-e^{-y}cosx$

Differentiate w.r.t x

Then, we get

$u_x=-e^{-x}cosy+e^{-y}sinx$

Again differentiate w.r.t x

$u_{xx}=e^{-x}cosy+e^{-y}cosx$

Now differentiate u w.r.t y

$u_y=-e^{-x}siny+e^{-y}cosx$

Again differentiate w.r.t y

$u_{yy}=-e^{-x}cosy-e^{-y}cosx$

Substitute the values in given Laplace's equation

$e^{-x}cosy+e^{-y}cosx-e^{-x}cosy-e^{-y}cosx=0$

Hence, given function is a solution of given Laplace's equation.

2. $u=sinx coshy+cosx sinhy$

Differentiate w.r.t x

$u_x=cosx coshy-sinx sinhy$

Again differentiate w.r.t x

$u_{xx}=-sin x coshy-cosxsinhy$

Now, differentiate u w.r.t y

$u_y=sinx sinhy+cosx coshy$

Again differentiate w.r.t y

$u_{yy}=sinx coshy+cosx sinhy$

Substitute the values then we get

$-sinx coshy-cosxsinhy+sinxcoshy+cosx sinhy=0$

Hence, given function is a solution of given Laplace's equation.

4 0

2 years ago

Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at

DiKsa [7]

Salt flows in at a rate of (5 g/L)*(3 L/min) = 15 g/min.

Salt flows out at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

So the net flow rate of salt, given by $x(t)$ in grams, is governed by the differential equation,

$x'(t)=15-\dfrac{3x(t)}{10}$

which is linear. Move the $x$ term to the right side, then multiply both sides by $e^{3t/10}$ :

$e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}$

$\implies\left(e^{3t/10}x\right)'=15e^{3t/10}$

Integrate both sides, then solve for $x$ :

$e^{3t/10}x=50e^{3t/10}+C$

$\implies x(t)=50+Ce^{-3t/10}$

Since the tank starts with 5 g of salt at time $t=0$ , we have

$5=50+C\implies C=-45$

$\implies\boxed{x(t)=50-45e^{-3t/10}}$

The time it takes for the tank to hold 20 g of salt is $t$ such that

$20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}$

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2 years ago

)A mule deer can run one over four of a mile in 25 seconds. At this rate, which expression can be used to determine how fast a m

makvit [3.9K]

Well, if you take x to be 25 seconds, there are 2.4x in 1min. So, that multiplied by 60, would be 144x. Therefore, 1/4 multiplied by 144x should be the answer.

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2 years ago

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Simon has 160160160 meters of fencing to build a rectangular garden.

yaroslaw [1]

Answer:

25

Step-by-step explanation:

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1 year ago

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3) You find a jar of quarters on the sidewalk and decide to start collecting them to cash in at the end of the school year.

soldier1979 [14.2K]

Answer:

I dont rally know

Step-by-step explanation:

try it yourself

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1 year ago