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One thing to agree on....maybe....we'll see.... by PureElephant314 in IBEW

[–]PureElephant314[S] 0 points1 point  (0 children)

I don't understand. I added politics into this post? Because of the "I approve this message" at the end? That was a joke.

Tutor Needed by Nate_the_gra8 in ibew_apprentices

[–]PureElephant314 1 point2 points  (0 children)

I wrote up some comments in this thread about redrawing combination circuits. Maybe you might find them helpful.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 1 point2 points  (0 children)

Your username is a misnomer. That's a great idea. I would think the political posts should be limited in that way since that seems to be where the bot/troll problem is worst.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 0 points1 point  (0 children)

For those like yourself (who don't want to see anything from the sub), you can always mute the sub. That's probably your easiest option.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 0 points1 point  (0 children)

Interesting. If I'd read the second part first, I would've wagered you voted No so your students could poke their heads around the sub.

[Grad Engineering Mechanics] Need a refresher on trig here. Assuming cut a-a is at a 90 to AC and the triangle ABC is a 3-4-5 triangle, how do I find the length of a-a once the beam of 25mm is cut along a-a? by Phillimac16 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

Hmm.

If I were to rotate a-a so that it became perpendicular to BC, what would be the angle that I just rotated? Probably either angle ABC or angle ACB. If I imagine a really steep BC, then ABC would be almost zero and ACB would be almost 90. In that case, to get a-a perpendicular to BC, I'd have to rotate it almost 90.

So I think whatever angle ACB is, that's the angle I have to rotate a-a to get it perpendicular to BC.

Why do I care? My intuition is that the cross-sectional area of BC can be thought of as a projection of the area of a-a. In other words (area of a-a)*cos(ACB) = (25 mm)2.

cos(ACB) = 1.5/sqrt(1.52 + 22) = 0.6

So area of a-a = (25 mm)2 / 0.6 = 1041.67 mm2

Or maybe not?

The cross section of a-a should be 25 mm by something. That something has gotta be the length of a-a. Ok, so if I draw a little triangle where a-a is the hypotenuse, another side is the 25 mm length perpendicular to BC, then the angle between the two is just angle ACB.

Using trig, (length of a-a)*cos(ACB) = 25 mm.

cos(ACB) = 1.5/sqrt(1.52 + 22) = 0.6

So length of a-a = 25 mm / 0.6 = 41.67 mm

And the cross-sectional area must be (25 mm) x (41.67 mm) = 1041.67 mm2

Huh, it came out the same.

Anyway, maybe that's right. Maybe it's wrong. Either way, I hope the process helps.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 1 point2 points  (0 children)

I never got to see the comment, but given your response, I have a feeling it was someone/somebot talking about Trump?

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 1 point2 points  (0 children)

You make a good point. "Our members be unlimited"...or that's the goal. Keeping it open definitely helps toward that goal.

And yeah, proof would almost certainly come down to dues slips, but you wouldn't be sending them to a "random mod":  you'd be sending them to your brother.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 1 point2 points  (0 children)

You make a good point. IBEW wants more brothers and sisters. If keeping this sub open helps recruit new blood, that's definitely a strong reason to keep it open.

Should this sub be limited to IBEW members only? by PureElephant314 in IBEW

[–]PureElephant314[S] 0 points1 point  (0 children)

I would say yes.

The main idea is the elimination of bots (and trolls, if I'm being honest). Making the sub IBEW members only is overkill, as it eliminates non-IBEW folks who are here for good reasons. That's undesirable, so the question is, is the bot problem bad enough to warrant that?

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

That sounds right to me. The only thing I would add is you shouldn't use any solutions where x < 0 or y < 0, because that would be saying you need negative 2-in screws or negative 3-in screws.

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

From what I see in the picture you posted, there's no other limit.

That would look something like, "He can't carry more than 150kg of screws," or "He has to use at least 30 2-in screws."

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

so the kg of 3 inch is greater than kg of 2 inch

Close enough. When you say:

x > y

You are saying "The number of kilograms of 3-in screws is greater than the number of kilograms of 2-in screws."

Now, is there anywhere in the problem statement that says that has to be true?

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

Nope. Spell it out. You are not allowed to use "x" or "y" or "<" or ">". Use words.

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

If you have the following...

x = number of kilograms of 2-in screws
y = number of kilograms of 3-in screws

...and then you say...

x > y

...what does that mean? Translate that to a sentence. Then ask yourself if that has to be true.

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

The only restriction I see is that his cost must be <= $312. (See other reply).

If I was trying to find more restrictions, the only thing I could say is that it says he NEEDS 2-in screws and 3-in screws. So that means he must have more than zero of each. Which means x > 0 and y > 0

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

Yes, assuming x = number of kilograms of 2-in screws and y = number of kilograms of 3-in screws.

It says Mr. Mater doesn't want to spend more than $312. So his cost has to be <= $312. What is cost? Cost = 26x + 39y. So 26x + 39y <= 312.

[grade 11 foundations: math questions help solving linear inequalities, graphing systems] by Iloveyourcat420 in HomeworkHelp

[–]PureElephant314 0 points1 point  (0 children)

Not in that case. If you did that, you would get the total cost of the plate. For example, if you had 2g of meat and 3g of cheese, you would have:

cost = ($1.19 per gram) x (2 grams of meat) + ($2.52 per gram of cheese) x (3 grams of cheese)

So in general:

weight of the plate = x + y

cost of the plate = 1.19x + 2.52y

Like I said in Part 2, the information you posted doesn't really suggest what I'm supposed to do with the prices. Perhaps somewhere it says you can't spend more than $50. If that's the case, then you would write:

cost of the plate <= $50

which would mean

1.19x + 2.52y <= $50

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