Answer:
5, -5
Step-by-step explanation:
well 5-5 is 0 so that's it
:))))
Answer:
the answer that 9-(2,4)
Step-by-step explanation:
We have that an inverse variation in a function is given by:
We must find the value of the constant k.
To do this, we substitute the values of x1 = 12, y1 = 15.
We have then:
Clearing k we have:
Substituting the value of k we have that the inverse variation is:
For y2 = 3 we have:
Clearing x2 we have:
Answer:
Answer:
M,L,K
Step-by-step explanation:
The angle opposite the side is in the order of their size. So the larger the side the larger the angle
<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>
