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Advocard [28]
2 years ago
8

Show that the sum of two irrational numbers can be rational ​

Mathematics
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

Hint: We will have to know about the rational numbers and irrational numbers. ... So, the sum of the given two irrational numbers is equal to 6 which is a rational number in the form of p/q where p=6 and q=1 both are integers. Therefore, it is proved that the sum of the two given irrational numbers is a rational number.

Step-by-step explanation:

Hope it helps u

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Write a quadratic equation ax²+ bx + C = 0, whose coefficients are a =-9, b = 0, c = -2
aksik [14]
<h3>ax² + bx + c = 0</h3>

<em>Let's write -9 where we see A</em><em>:</em>

<h3>-9x² + bx + c = 0</h3>

<em>Let's</em><em> </em><em>write</em><em> </em><em>0</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>B</em><em>:</em>

<h3>-9x² + 0.x + c = 0</h3>

<em>(</em><em>Since B = 0, when it is multiplied by x, it becomes 0 again</em><em>)</em>

<h3>-9x² + c = 0</h3>

<em>Let's</em><em> </em><em>write</em><em> </em><em>-2</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>C</em><em>:</em>

<h3>-9x² + -2 = 0</h3>

<em>Now we can move on to solving our equation</em><em>:</em><em>)</em>

<em>Let's put the known and the unknown on different sides:</em>

<em>(</em><em>-2 goes to the opposite side positively</em><em>)</em>

<h3>-9x² = 2</h3>

<em>(</em><em>i</em><em>t goes as a division because it is in the case of multiplying -9 across</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<h3>x² = 2/-9</h3>

<em>I could not find the rest of it, but I did not want to delete it for trying very hard. Sorry. It felt like we should take the square root, but I couldn't find it, maybe this can help you a little bit.</em>

<em>Please do not report</em><em>:</em><em>(</em>

<em>I hope I got it right, I'm trying to improve my English a little :)</em>

<h3><em>Greetings from Turke</em><em>y</em><em>:</em><em>)</em></h3>

<h3><em><u>#XBadeX</u></em></h3>
8 0
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