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GREYUIT [131]
3 years ago
11

Is 0.010010001..... rational or irrational

Mathematics
1 answer:
zlopas [31]3 years ago
3 0

0.010010001..... is an irrational number since it keeps repeating. Rational numbers are mostly whole numbers that don't repeat forever.

Your answer is irrational.

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Eduardwww [97]
The "e" just lets you know that it's scientific notation.
5 0
3 years ago
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there are 7209 students at a amusement park. to the nearest thousands, how many students are in the park
Karo-lina-s [1.5K]

Step-by-step explanation:

In ordee to round the number 7209 to the nearest thousands, we take a look at the digit in the hundreds place (which is 2).

Since 2 < 5, we round the number itself down to the nearest thousands:

7209 => 7000.

Hence our rounded number is 7000.

4 0
4 years ago
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Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)
Drupady [299]

Answer:

The equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

Let us solve the expression step by step to get the equivalent

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

as

\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}      ∵ \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=x^{2\cdot \frac{1}{7}}

=x^{\frac{2}{7}}

also

\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}         ∵  \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=y^{3\cdot \frac{1}{5}}

=y^{\frac{3}{5}}

so the expression becomes

\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)            ∵ \:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}

Thus, the equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

5 0
3 years ago
Plllllllllleeeeeeeeeeeesssse help!
ryzh [129]

Answer:

B) y = 3x + 2

Step-by-step explanation:

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5 0
3 years ago
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Which of the following represents the geometric sequence?
anastassius [24]
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----------------------------------

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Choice C is not the answer either since most of the terms are positive but the -4 in slot 3 is not positive. If it was positive, then the common ratio would be 4 and this sequence would be geometric. If the terms alternated between positive and negative, then it would work as well (r would be r = -4)

Choice D has alternating terms of positive and negative, but there's no change in absolute value. There is no growth or decay as this the terms effectively stay the same. So there is no common ratio. This is why we can rule out choice D.
4 0
4 years ago
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