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3 years ago

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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Select more than 1 please

umka2103 [35]

Here are the answers:
1.) DC=D’C’
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3 years ago

Need help on this one please

Archy [21]

The answer is A because there are only acute angles

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3 years ago

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