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TiliK225 [7]
4 years ago
7

Which is greater 0.01 or 0.1 or are they equal?

Mathematics
2 answers:
Fudgin [204]4 years ago
8 0

Answer:

Greater

Step-by-step explanation:

0.01 is Greater. Because it's the 100th part of 1. Whereas 0.001 is 1000th part of 1. For example, a bread divided in 100 parts will have bigger pieces.

Drupady [299]4 years ago
6 0
0.1 is the greater number
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Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4
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<h3><em>The complete question:</em></h3>

<u><em> </em></u><u>Harold uses the binomial theorem to expand the binomial </u>(3x^5 -\dfrac{1}{9}y^3)^4<u />

<u>(a)    What is the sum in summation notation that he uses to express the expansion? </u>

<u>(b)    Write the simplified terms of the expansion.</u>

Answer:

(a). (3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$

(b).(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}

Step-by-step explanation:

(a).

The binomial theorem says

(x+y)^n=$$\sum_{k=0}^{n}  \binom{n}{k}x^{n-k}y^k $$

For our binomial this gives

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}x^{4-k}y^k $$}

(b).

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$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4

$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}   }

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