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2 years ago

Is 17/2 rational or irrational?

Mathematics

2 answers:

Naya [18.7K]2 years ago

8 0

Answer:

it's a Rational number

Step-by-step explanation:

because it repeats and stops

Katyanochek1 [597]2 years ago

5 0

Hi, I'm happy to help!

A number is rational if it repeats itself, or stops. This number, both in fraction and decimal form (8.5), stops. This means that the number is rational.

I hope this was helpful, keep learning! :D

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Triangle EFG has the following vertices E(1,2), F(4,2) and G(1,6). What are the new coordinates of triangle E’F’G’ after it is r

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Answer:

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2 years ago

The average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00. Assuming a normal distribution, if 77 percent

jek_recluse [69]

Answer:

22.97

Step-by-step explanation:

Given that the average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00.

Let x be the average cost an IRS Form 1040 tax filing at Thetis Tax Service

$P(X$ is given

From std normal distribution table we find 77th percentile z value.

z=0.74

Corresponding X value = 155

i.e. $0.74=\frac{155-138}{s}$ where s is the std deviation

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2 years ago

Gail bought four snacks that cost $1.00, $1.50, $2.50, and $3.00. What is the average of the deviations from the mean?

Deffense [45]

The answer for this is 2

3 0

2 years ago

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Measure the screwdriver's length to the nearest tenth of a centimeter. Express the answer as a decimal.

Alja [10]

Answer:

17.1 cm

Step-by-step explanation:

A screw driver is a mechanical tool or device which is mainly used for screwing the screws and unscrewing them. It is also used for removing the nuts and bolts and also serves a variety of uses.

It is typically made of steel and has a handle and a shaft which ends as a tip.

In the context, the length of the screw driver from the given figure expressed to the nearest tenth of a centimeter is 17.1 cm.

It is also equivalent to $6\frac{3}{4}$ inch.

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3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

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3 years ago