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3 years ago

13 is the same as 1÷3. What is the exact quotient in decimal form of 1÷3?

Mathematics

2 answers:

elixir [45]3 years ago

7 0

Answer: Hence, Quotient of 1÷3 is given by

1) 0

2) 0.333333....

Step-by-step explanation:

Since we have given that $\frac{1}{3}=1\div 3$

We need to find the exact quotient in decimal form :

$\frac{1}{3}=0.33333................$

Exact quotient will be '0' only.

Exact quotient in decimal is 0.333333333......

Hence, Quotient of 1÷3 is given by

1) 0

2) 0.333333.....

Yuliya22 [10]3 years ago

6 0

Answer:

Exact quotient is 0

Step-by-step explanation:

Given : $\frac{1}{3}$ is the same as 1÷3.

To find : What is the exact quotient in decimal form of 1÷3

Solution : Here the divisor is 3

dividend is 1

To find quotient we divide divisor by dividend

1÷3 or $\frac{1}{3}$ = 0.333...

So, the quotient is 0.333... and remainder goes on to 1 till the division goes on.

Exact quotient in decimal form is 0

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Read 2 more answers

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klio [65]

It looks like the differential equation is

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

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

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

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

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 :

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