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#1 Re: Help Me ! » complex numbers » 2016-05-17 16:46:50
The j is a replacement of the regular i.
j is basically the same as if we were to say i and both indicate (square root negative 1)
I think j's are likely used when performing problems that are in relation to electricity.
j = square root negative 1
i = square root negative 1
#2 Re: Help Me ! » Modular arithmetic » 2016-05-17 09:41:44
From A - L, I am guessing that they would be 0, 1, 2, 3, 4, . . . 8.
There's a noticeable pattern when we look at the other two charts and they also go as 0, 1, 2, 3, 4, . . . 8.
Although there are different variables found upon the charts, the third one has to be similar to the other two.
For example, it could be like at one second there is a voltage of 2, at the 2nd second theres a voltage of 19, at the 3rd second theres a voltage of 14 . . .
These charts could represent three different individual testings on the voltages found upon three different electric producing products.
Primarily, I am using this as an example, where these charts could be used in real world application.
Does this help?
#3 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 20:23:19
So . . .
5 square root 3 times square root 2 / square root 2 times square root 2
5 square root 6 / square root 4
5 square root 6 / 2! 
Thanks for helping me solve this lengthy problem bobbym!
#4 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 20:11:57
Awesome! So now that we have found h = square root (75/2). How would the process of simplifying work out? I am a bit confused when approaching this step.
I continued forth by simplifying the square root 75 into square root 25 and square root 3. The square root 25 became 5 and so I came up with an answer of (5 square root 3 / square root two). However, the final answer is supposed to be 5 square root 6 / 2.
#5 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 19:57:14
Yes! I follow through by canceling out the square roots of the two's and then multiplying the numerators by themselves. Thus, we obtain the new numbers of 25/2 and 100/2. 100/2 = 50. Subtract both sides by 25/2. h^2 = 50 - (25/2). h^2 = 75/2. h = square root (75 / 2) or square root 75 / square root 2. Am I correct so far?
(I should have left the 100/2 as it was)
#6 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 19:39:33
Well that explains quite a bit! Would it be correct to subtract (5 / square root 2)^2 from both sides of your previous equation, and come up with h^2 = (10 / square root 2)^2 - (5 / square root 2)^2 to obtain the height of the triangle? And if so, how would I continue with solving it? I have difficulties at this point.
Also, if the sides of an equilateral triangle were 20 / square root 2 and I halved it, would the new base of the 30-60-90 triangles be 10 / square root 2?
(Just checking to see if I have a full understanding of this)
#7 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 19:22:52
How did you get the 5 / square root 2? Did you just divide 10 / square root 2 by 2?
#8 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 19:10:32
How the height and (new) base were obtained step by step.
(Primarily the height)
#9 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 19:03:39
I know that the answers to this question are: a height of (5 square root 6 / 2). However, I don't quite understand how this answer was obtained. There isn't much of an explanation in the book I have.
Also, I am in a section where everything is based solely on geometry, so I find that it wouldn't be quite necessary to use socahtoah as it would be more suitable for a trig problem. I know that the Pythagorean theorem is one of the options I can choose in solving this problem, however, I only have difficulties with solving it because of having a square root involved in the problem.
#10 Re: Help Me ! » Finding the height of an equilateral triangle » 2016-05-16 18:53:51
If the original sides of an equilateral triangle is (10/ square root 2) and I divide the equilateral triangle into halves, would I divide
(10/ square root 2) / 2 to obtain the new length for one of the sides in the 30-60-90 triangle (the side opposite of the 30 degree angle)? Then continue on with the Pythagorean theorem?
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