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3 years ago

Men can reap a field in 20 days, in how many days will 15 men reap the field?

Mathematics

1 answer:

Hitman42 [59]3 years ago

4 0

1.333333333333333333333333333333

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The given function k is a composition of two functions m

lara31 [8.8K]

Answer:2/4 √x (2 over 4 square root x)

Step-by-step explanation:

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3 years ago

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Someone help me plz hurry -51/7

Maurinko [17]

Answer:

-51/7

Step-by-step explanation:

-51/7 cannot be simplified

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3 years ago

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What is the Area of 28in circle rounded to the nearest hundred

Slav-nsk [51]

Answer:

The circumference is similar to the perimeter in that it is the total length needed to draw the circle.

We note the circumference as c.

c = 2πr

c = πd

This depends on whether or not you know the radius (r) or the diameter (d)

Let’s calculate one manually, for example.

If r = 6 cm, then the circumference is c = 2π(6) = 12π cm, if writing in terms of π. If you prefer a numerical value, the answer rounded to the nearest tenth is 37.7 cm.

Suppose you only know the diameter? If the diameter is 8 cm, then the circumference is c = π(8) = 8π or 25.1 cm, rounded to the nearest tenth.

Step-by-step explanation:

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3 years ago

Can anyone please work out question 10 in the picture and clearly explain it? Would appreciate it

kotykmax [81]

9945/ 39= 255
If you were to try to divide 99 by thirty five it would not be a whole. You would be able to multiply 39 by 2 to get 78. 99 subtracted by 78 = 21 you would then carry down the 4. to equal 214 you would then figure out 39 multiplied by 5 is the highest number to figure a 195 you would have 19 left over. Carry down your 5 and would have 195. You would have 195 again and then multiply 39 by 5 and it would equal out to 195. Your answer would then be 255.

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3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$