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Ber [7]
3 years ago
8

What is the area of a rectangle with sides 10cm and 5cm

Mathematics
2 answers:
ratelena [41]3 years ago
7 0
You multiply 10 by 5 to get 50. So your answer is 50cm.
velikii [3]3 years ago
6 0
The answer is 50cm^2
because u have to multiply 10 times 5 equals 50
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If pi is infinite, how does it work? How do we keep finding new numbers?? What is pi?
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3.141592653589793238

Step-by-step explanation:

Succinctly, pi—which is written as the Greek letter for p, or π—is the ratio of the circumference of any circle to the diameter of that circle. If you divide the circumference of the circle by the diameter, you will get approximately 3.14—no matter what size circle you drew! A larger circle will have a larger circumference and a larger radius, but the ratio will always be the same. If you could measure and divide perfectly, you would get 3.141592653589793238..., or pi.

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2 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
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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

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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.

7 0
3 years ago
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