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Posted
2 hours ago, RIDDLE said:

how to do question 28-30

 

 

1)

Each of questions 28, 29 and 390 should have their own thread.

2) Have you had no thoughts about any of these to tell us?

3) Taking the average question, since it is the simplest, what would you do in elementary algebra if faced with the question

Simplify


[math]\frac{{a + b}}{{a - b}}[/math]


Simplify means write  out the expression on one line not two.

  • 3 weeks later...
Posted

Studiot you have me intrigued as this is basic enough math that I should be able to handle it. I dont want to blow it for the OP and actually dont know I'm right.

But I start by multiplying by -1/-1

Is it the right direction?

Posted (edited)
2 hours ago, druS said:

Studiot you have me intrigued as this is basic enough math that I should be able to handle it. I dont want to blow it for the OP and actually dont know I'm right.

But I start by multiplying by -1/-1

Is it the right direction?

Yes, multiply by 1 (why the minus signs?)

Now 1 equals


[math]1 = \frac{{\left( {1 + i} \right)}}{{\left( {1 + i} \right)}}[/math]

 

So we have


[math]\frac{{\left( {1 + i} \right)}}{{\left( {1 - i} \right)}}*\frac{{\left( {1 + i} \right)}}{{\left( {1 + i} \right)}} = \frac{{\left( {1 + 2i + {i^2}} \right)}}{{{{\left( 1 \right)}^2} - {{\left( i \right)}^2}}} = \frac{{2i}}{2} = i[/math]

Since the bottom then becomes the difference of two squares.

I think this thread is now old enough that the OP has lost interest.

Edited by studiot
  • 1 month later...
Posted

28 is straight forward if you know some basic facts:

The midpoint of the line segment between (x_1, y_1) and (x_2, y_2) is ((x_1+ x_2)/2, (y_1+ y_2)/2).  The slope of the line between those two points is m= (y_2- y_1)/(x_2- x_1).  The line perpendicular to y= mx+ b has slope -1/m.  The distance between points (x_1, y_1) and (x_2, y_2) is sqrt((x_2- x_1)^2+ (y_2- y_1)^2).

Whoever gave you this problem expects you to know that!

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