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3 years ago

Canh square roots be rational numbers ?

2 answers:

8 0

When the square root of a number is a whole number, this number is called a perfect square. ... Not all square roots are whole numbers. Many square roots are irrational numbers, meaning there is no rational number equivalent

irinina [24]3 years ago

7 0

Very good question, i’ll try to answer this as best as I can so here we go hope you enjoy

All rational numbers have the fraction form
,
a
b
,
where a and b are integers(≠0
b
≠
0
).

My question is: for what
a
and
b
does the fraction have rational square root? The simple answer would be when both are perfect squares, but if two perfect squares are multiplied by a common integer
n
, the result may not be two perfect squares. Like:
49→818
4
9
→
8
18

And intuitively, without factoring, =8
a
=
8
and =18
b
=
18
must qualify by some standard to have a rational square root.

Once this is solved, can this be extended to any degree of roots? Like for what
a
and
b
does the fraction have rational
n
th root?

Have a blessed day God bless.

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What polyhedron can be formed from a net diagram that has 2 congruent hexagons and 6 congruent rectangles?

Harman [31]

Answer A. Hexagonal prism.

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3 years ago

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3 years ago

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Find a1, for the arithmetic series with S14 = - 420 and d = -6.

faust18 [17]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} n = 14\\ d= -6 \end{cases} \\\\\\ a_{14}=a_1+(14-1)(-6)\implies a_{14}=a_1+(13)(-6)\implies a_{14}=a_1-78 \\\\[-0.35em] ~\dotfill$

$\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ \cline{1-1} n= 14\\ S_{14}=-420\\ a_{14}=a_1-78 \end{cases}\implies S_{14}=\cfrac{n(a_1+a_{14})}{2} \\\\\\ -420=\cfrac{14[a_1+(a_1-78)]}{2}\implies -420=7(2a_1-78)\implies \cfrac{-420}{7}=2a_1-78 \\\\\\ -60=2a_1-78\implies 18=2a_1\implies \cfrac{18}{2}=a_1\implies 9=a_1$

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3 years ago

How do I simplify 9v+9(1+7v) and can you show me the work

KonstantinChe [14]

Answer:72v+9

Step-by-step explanation:

Apply the distributive property.

9 v + 9 ⋅ 1 + 9 ( 7 v )

Multiply

9 by 1 . 9 v + 9 + 9 ( 7 v )

Multiply

7 by 9 . 9 v + 9 + 63 v

Add

9 v and 63 v . 72 v + 9

I hope this helps you.

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3 years ago

) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

Let $p_1$ = population proportion of drivers 16-24 years old who buckle up .

$p_2$ = population proportion of drivers 25-69 years old who buckle up .

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, test statistics = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a difference in seat belt use.

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3 years ago