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lys-0071 [83]
3 years ago
8

Is 0.62 a rational number?

Mathematics
2 answers:
dimaraw [331]3 years ago
7 0

A rational number is any number that can be written <em>as a ratio of two integers</em>.

0.62 fits the bill, as it can be written as the ratio 62/100 (or simplified to the ratio 31/50)

Bingel [31]3 years ago
4 0
Yes man yes man yes yes
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Which of the following does NOT represent a function ?
Veseljchak [2.6K]

Answer:. C is not a function.

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8 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
You have 1 case of soap bars and there are 150 bars in a case. You use 300 bars of soap per day. how many cases do you news to o
nignag [31]
You would need 13 cases. If you would like an explanation of my answer, let me know.
7 0
3 years ago
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