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VladimirAG [237]
3 years ago
8

What is the simplified form of square root of 144x^36?

Mathematics
2 answers:
Anna11 [10]3 years ago
6 0

Answer:

Option (b) is correct.

Expression becomes =12x^{18}

Step-by-step explanation:

Given : square root of 144x^{36}

We have to write the given expression in simplified form.

Consider the given expression square root of 144x^{36}

Mathematically written as \sqrt{144x^{36}}

\mathrm{Apply\:radical\:rule\:}\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},

We have,

=\sqrt{144}\sqrt{x^{36}}

We know \sqrt{144}=12

=12\sqrt{x^{36}}

\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0

\sqrt{x^{36}}=x^{\frac{36}{2}}=x^{18}

Thus, Expression becomes =12x^{18}

trasher [3.6K]3 years ago
3 0

Answer:

The correct option is B.

Step-by-step explanation:

The given expression is

\sqrt{144x^{36}}

Use the property of radicals.

\sqrt{ab}=\sqrt{a}\sqrt{b}

\sqrt{144x^{36}}=\sqrt{144}\times \sqrt{x^{36}}

Use the property of exponent.

a^{mn}=(a^m)^n

\sqrt{144x^{36}}=12\times \sqrt{(x^{18})^2}

\sqrt{144x^{36}}=12\times x^{18}

\sqrt{144x^{36}}=12x^{18}

Therefore option B is correct.

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The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a s
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For this case we can solve this problem creating the following table

Number of particles      Frequency         Rel. Frequency

             0                            1                       1/100 =0.01

             1                             2                      2/100 =0.02

             2                            3                       3/100=0.03

             3                            12                     12/100=0.12    

             4                            11                       11/100=0.11

             5                            15                      15/100=0.15

             6                            18                      18/100=0.18

             7                            10                       10/100=0.1

             8                            12                       12/100=0.12

             9                            4                         4/100=0.04

             10                           5                         5/100=0.05

             11                            3                         3/100=0.03

             12                           1                          1/100=0.01

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             14                           1                          1/100=0.01

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P(X\geq 1) = 1-P(X

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P(X\geq 5) = 1-P(X

And replacing we got:

P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71

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