anonymous
  • anonymous
Short Leg= 2x-6 Long leg= 2x Longest= 30 Find the missing sides to get 30
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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tkhunny
  • tkhunny
Perhaps there was a figure or a description of a figure?
anonymous
  • anonymous
I can't screenshot the figure. The codes in the page prevent it.
tkhunny
  • tkhunny
And you are entirely unable to describe it?

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More answers

anonymous
  • anonymous
its Pythagoreon Theorem
anonymous
  • anonymous
a squared plus b squared equals c squared
tkhunny
  • tkhunny
Okay, then you should apply that. Can you identify a, b, and c?
anonymous
  • anonymous
I did.
anonymous
  • anonymous
Reread the question.
tkhunny
  • tkhunny
You did what? Fill n the blanks... For the Pythagorean Theorem, I need to identify a, b, and c. a = ?? b = ?? c = 30
anonymous
  • anonymous
I did.
anonymous
  • anonymous
A equals short leg
anonymous
  • anonymous
B equals the other leg
anonymous
  • anonymous
C equals longest leg
tkhunny
  • tkhunny
Okay. What did the Pythagorean Theorem look like when you substituted those values?
anonymous
  • anonymous
When I squared them all?
tkhunny
  • tkhunny
Did you do that? Yes. What did you get?
anonymous
  • anonymous
I have not done it yet because that and the whole problem is where my confusion lies.
tkhunny
  • tkhunny
I think we have established this: a = 2x-6 b = 2x c = 30 We have also established the Pythagorean Theorem \(a^{2} + b^{2} = c^{2}\) Now our task is to substitute the given values into the Pythagorean Theorem. You do that and let's see if you wander off.
anonymous
  • anonymous
I wander off in the beginning.
tkhunny
  • tkhunny
Stop saying that and simply start writing. Change your screen name. Do SOMETHING so that you can win one battle, here. You seem to be in the wrong class. How can we move you toward a more positive outcome? I can't get you to write ANYTHING. That's not good.
anonymous
  • anonymous
I have Aspbergers. Either bear with my incompetence that I was born with or help me find someone who can help.
tkhunny
  • tkhunny
You didn't answer my question. Can you substitute the known values into the known formula, or not? Do you know what substitution means? Have you ever seen the process?
anonymous
  • anonymous
Thank you for your help. Good day.
anonymous
  • anonymous
What tkhunny is asking is which is the hypothenuse and the adjacent side.
anonymous
  • anonymous
What tk hunny was not understanding is I wrote them already and as neither side is in normal form I am confused.
anonymous
  • anonymous
Would you be able to use the draw option under the chat log to drwa the picture for us?
anonymous
  • anonymous
maybe im on a laptop
anonymous
  • anonymous
Just do your best, if you have problem explaining it is much easier to see a picture. If we can't understand your question, we will not be able to help.
anonymous
  • anonymous
|dw:1354487589391:dw|
anonymous
  • anonymous
okay tell me if im right:|dw:1354487622965:dw|
anonymous
  • anonymous
swap the legs
anonymous
  • anonymous
the one that are perpendicular?
anonymous
  • anonymous
2x should be on the bottom with 2x minus 6 on the left the 30 is in the right spot
anonymous
  • anonymous
|dw:1354487934867:dw|
anonymous
  • anonymous
okay that helps, so you want to solve for x. Because you have 1 variable ( which in this case would be x, and I'm assuming you know what variable means) In this case you have a formula relating all angle. We will call it as follow ( because a square + b square = c sqare is too vague and can confuse you): Opposite + adjacent = hypothenuse. Now you're hypothenuse is a mathematical term for what you called " longest leg" which is mathematically wrong to say. Your opposite and adjacent is the two other "Legs" and is relative (therefore meaning it all depends on how you see it) to what you want. I relate the BOTTOM RIGHT angle always. THerefore my opposite side is the 2x-6 and the adjacent is 2x Now come the equation which is as follow: adjacent(squared) + opposite(squared) = hypothenuse (squared) Plug in ( therefore replace the word corresponding to the right side) with the right equation, and solve for x Can you solve for x?
anonymous
  • anonymous
I get legs. I don't get math terms. Can you show me a visual?
anonymous
  • anonymous
|dw:1354488141269:dw|
anonymous
  • anonymous
o.O
anonymous
  • anonymous
does this help?
anonymous
  • anonymous
No.
anonymous
  • anonymous
yikes. Hum alright, ill try another way. When you solve for an equation, at your level, you have only one variable : x
anonymous
  • anonymous
Ok stop just for a moment. You are going way too bloody complex. I don't get complex and I don't get abstract.
anonymous
  • anonymous
Ah I'm sorry my friend then, I hope you find someone who can give you a better explanation
anonymous
  • anonymous
I have Aspbergers I am used to not being able to find help really.
tkhunny
  • tkhunny
Why as Aspergers a restriction on your ability to visualize, follow abstraction, or communicate symbolically? It that the only diagnosis? You are so SURE that you CAN'T get it, and you know EXACTLY what it is you can't get, that I have to wonder if we don't have other pathologies at play, here. If you don't get abstract, complex, or math symbols, how do you know you have encountered such a thing?
anonymous
  • anonymous
Have you ever looked up Aspbergers? Researched it? Lived it?
anonymous
  • anonymous
By your replies I can say no you have not. Aspbergers is a Social Disorder in the DMV book and can affect the ability to comprehend abstract concepts in Math, English, Reading, Writing, and also affects Social Skills.
tkhunny
  • tkhunny
Yes, Yes, and Yes. But I had no cognitive difficulty related to mathematics. That's why I'm puzzled. It's quite a different presentation than I've previously encountered. I've only seen social aspects.
anonymous
  • anonymous
Not every Aspbergers person is the same.
tkhunny
  • tkhunny
Clearly. Are you aware of what level of abstration can you handle? You seemed to be okay calling a leg of a triangle '2x' and another leg '2x-6'.
anonymous
  • anonymous
It is obvious you can't help me. Good day.
tkhunny
  • tkhunny
Not if you don't talk to me and you just punt.
anonymous
  • anonymous
New problem got it wrong :( let me write the new one
anonymous
  • anonymous
nvm same problem as this one lol
mathstudent55
  • mathstudent55
short leg = 2x - 6 long leg = 2x longest side = 30 Pythagorean theorem: short leg squared + long leg squared = longest side squared (2x - 6)^2 + (2x)^2 = 30^2 (2x - 6)(2x - 6) + 4x^2 = 900 4x^2 - 12x - 12x + 36 + 4x^2 = 900 8x^2 -24x - 864 = 0 x^2 - 3x - 108 = 0 (x - 12)(x + 9) = 0 x - 12 = 0 or x + 9 = 0 x = 12 or x = -9 Since long leg is 2x, 2(-9) = -18, and a leg can't have a negative length, discard the negative solution. x = 12 short leg: 2x - 6 = 2(12) - 6 = 18 long leg: 2x = 2(12) = 24 The lengths are 18 for the short leg, and 24 for the long leg.
anonymous
  • anonymous
Holy crap thank you. Been trying to get a straight answer on this problem for 3 hours. Thank you again.
anonymous
  • anonymous
Oh great word problems ugh hate those :/ will post that next
mathstudent55
  • mathstudent55
I figured you got all kinds of bs answers except a simple answer to your problem, so it was about time you got one.
anonymous
  • anonymous
I appreciate it greatly. Some just don't want to hear the word simple these days. To me simple is better and easier for me to understand. It is not I refuse to try I just don't always get it. Thank you again.
anonymous
  • anonymous
This question is closed sir.
anonymous
  • anonymous
To be simple is being fed the answer which I don't believe in. Even if you have Aspergers, Autism, I will never give the answer to somebody.
anonymous
  • anonymous
@MarcLeClair
anonymous
  • anonymous
Look sir your view is obviously different. I see simple as a way of understanding. Don't like it. Don't respond. Good day.
tkhunny
  • tkhunny
I see. You wanted someone to do the work for you, quite contrary to the forum rules. Is there ANY of that reply that you could not have produced? I couldn't get you to write a single symbol past the original post. Why? I'm pretty sure this forum is for learning, not for only answers to be copied and turned in as one's own work. Not a good day.

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