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Viefleur [7K]
3 years ago
13

Which of the following shows an integer and its opposite?

Mathematics
1 answer:
pashok25 [27]3 years ago
4 0

3, -3

Step-by-step explanation:

an integer is a whole number

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It can never be a square.
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For the pair of functions, find the indicated sum, difference, product, or quotient.
ycow [4]
For the pair of functions, find the indicated sum, difference, product, or quotient.

f(x) = 6 - 7x, g(x) = -3x + 7
Find (f + g)(x).

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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.

7 0
3 years ago
PLEASE HELP! I'LL GIVE BRAINLIEST!!!!!!!
daser333 [38]

Answer:

1.6x10^-18 gram

Step-by-step explanation:

(5.3 x 10⁻²³ gram/molecule) x (20,000 molecule)

= (5.3 x 10⁻²³ x 2 x 10⁴) gram

= (10.6 x 10⁻²³⁺⁴) gram

= (1.06 x 10⁻²³⁺⁵) gram

= 1.06 x 10⁻¹⁸ gram

We need to find the mass of 20,000 molecules of oxygen. It can be calculated using unitary method. Here, we can multiply 20,000 molecules  by the mass of one oxygen molecule.

7 0
2 years ago
Back to Content
Lelu [443]

Answer:

3

Step-by-step explanation:

3 0
2 years ago
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