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3 years ago

Is the difference between two rational numbers always rational number

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Not always. It depends on the teacher and the school. Depends on the rules and how your being tought.

Alenkinab [10]3 years ago

6 0

Hello,

An integer is a rational number (ex 5=5/1=10/2...)
The sum of 2 rationals is a rational.
Substract a rational is adding its opposite.
The difference of 2 rationals is always a rational (substraction is a intern law)

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2. You are given the hyperbolic relation modeled by xy = -2. Do the following: a) Rewrite the relation such that the dependent v

N76 [4]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The given equation is :

$xy = - 2$

1. The relationship such that dependent variable (y) is isolated is :

$y = \dfrac{ - 2}{x}$

2. The table accompanying this equation :

x = -4 and y = 0.5

x = -1 and y = 2

x = - 0.5 and y = 4

x = 0 and y = undefined

x = 0.5 and y = -4

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3 years ago

In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

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3 years ago

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zvonat [6]

Answer:

Step-by-step explanation:

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Find the distance between each pair of points. Round to the nearest tenth, if necessary.

Westkost [7]

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\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
O&({{ 0}}\quad ,&{{ 0}})\quad 
% (c,d)
P&({{ -5}}\quad ,&{{ 6}})\\
K&({{ 5}}\quad ,&{{ 0}})\quad 
% (c,d)
L&({{ -2}}\quad ,&{{ 1}})

\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
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3 years ago

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miss Akunina [59]

Answer:

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Step-by-step explanation:

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