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3 years ago

What is 1/8 divided by 2

Mathematics

2 answers:

erma4kov [3.2K]3 years ago

6 0

Answer:

Divide: 2 : 1

= 2

· 8

= 2 · 8

1 · 1

= 16

Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1

is 8

) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators.

In words - two divided by one eighth = sixteen.

SO FINAL ANSWER IS 16!!! HOPE THIS HELPS YOU mark me the brainliest if im correct

Afina-wow [57]3 years ago

4 0

0.0625. I even double checked.

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Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

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The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

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$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

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$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
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Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
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Integrate both sides with respect to $y$ to arrive at

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So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

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<img src="https://tex.z-dn.net/?f=%28%5Csqrt%282%29-%5Csqrt%28-2%29%29%28%5Csqrt%2818%29%2B%5Csqrt%28-2%29%29" id="TexFormula1"

prohojiy [21]

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Step-by-step explanation:

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