HOMEWORK help center ||HWH||open!

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Re: HOMEWORK help center ||HWH||open!

Postby Bobkitty246 » Mon Sep 14, 2015 11:25 am

O THANMYOU so much!!! <3
Now I get it lol
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Re: HOMEWORK help center ||HWH||open!

Postby Bobkitty246 » Mon Sep 14, 2015 11:31 am

TheFae wrote:
Bobkitty246 wrote:
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TYPE OF QUESTION:MATH
YOUR QUESTION:a line segment has one end point of A(6,-2) and a slope of -3/4. Find the coordinates of another possible endpoint B by adding the appropriate values to the coordinate of point A


PLEASE HELP ANYONE! I AM CONFUSED, AND UNSURE HOW TO SOLVE???!!!!! THANKYOU SO MUCH OMG


It's easier than you think. Take a deep breath!

So with the slope of -3/4, it means it is going down "3" points and over to the right "4" points. (rise over run is the same as y/x)
Now, adding the appropriate values would be the following:
y-coordinate: -2-3=-5
x- coordinate: 6+4=10.

Your new point is (10, -5). I hope this helped!


Wait...why would you plus the x and y values together? Doesn't that not allow?
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Re: HOMEWORK help center ||HWH||open!

Postby TheFae » Mon Sep 14, 2015 11:44 am

Bobkitty246 wrote:
TheFae wrote:
Bobkitty246 wrote:
Image
TYPE OF QUESTION:MATH
YOUR QUESTION:a line segment has one end point of A(6,-2) and a slope of -3/4. Find the coordinates of another possible endpoint B by adding the appropriate values to the coordinate of point A


PLEASE HELP ANYONE! I AM CONFUSED, AND UNSURE HOW TO SOLVE???!!!!! THANKYOU SO MUCH OMG


It's easier than you think. Take a deep breath!

So with the slope of -3/4, it means it is going down "3" points and over to the right "4" points. (rise over run is the same as y/x)
Now, adding the appropriate values would be the following:
y-coordinate: -2-3=-5
x- coordinate: 6+4=10.

Your new point is (10, -5). I hope this helped!


Wait...why would you plus the x and y values together? Doesn't that not allow?


I'm not adding the x and y values together, we're adding ONLY the 'y''s and ONLY the 'x's, which is allowed!




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Re: HOMEWORK help center ||HWH||open!

Postby Bobkitty246 » Mon Sep 14, 2015 12:23 pm

Thequote="Bobkitty246 wrote:
TheFae wrote:
Bobkitty246 wrote:
Image
TYPE OF QUESTION:MATH
YOUR QUESTION:a line segment has one end point of A(6,-2) and a slope of -3/4. Find the coordinates of another possible endpoint B by adding the appropriate values to the coordinate of point A


PLEASE HELP ANYONE! I AM CONFUSED, AND UNSURE HOW TO SOLVE???!!!!! THANKYOU SO MUCH OMG


It's easier than you think. Take a deep breath!

So with the slope of -3/4, it means it is going down "3" points and over to the right "4" points. (rise over run is the same as y/x)
Now, adding the appropriate values would be the following:
y-coordinate: -2-3=-5
x- coordinate: 6+4=10.

Your new point is (10, -5). I hope this helped!


Wait...why would you plus the x and y values together? Doesn't that not allow?


I'm not adding the x and y values together, we're adding ONLY the 'y''s and ONLY the 'x's, which is allowed![/quote]

Ah, yes YIKES. I died. Sorry. I'm stupid lol.
I keep forgetting that rise is Y, and run is X so MESSING myself up. Sorry thanks!
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Re: HOMEWORK help center ||HWH||open!

Postby sword » Tue Sep 15, 2015 11:06 am

Image
TYPE OF QUESTION: Math
YOUR QUESTION: If there's an area that's 10,000 x 10,000 feet on the surface, and 100 feet deep (so, the total volume of that would be calculated in cubic feet), how many tons would that be (volume, not weight)? I'm trying to calculate that because I need to know how much it would cost to mine an area that big for an assignment. I'm not really sure how to go about it (I'm tired and not thinking straight right now, lol).
Last edited by sword on Tue Sep 15, 2015 1:04 pm, edited 1 time in total.
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Re: HOMEWORK help center ||HWH||open!

Postby Waki » Tue Sep 15, 2015 12:49 pm

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TYPE OF QUESTION:(math
YOUR QUESTION:
Admiral wrote:If there's an area that's 10,000 x 10,000 feet on the surface, and 100 feet deep (so, the total volume of that would be calculated in cubic feet), how many tons would that be (volume, not weight)? I'm trying to calculate that because I need to know how much it would cost to mine an area that big for an assignment. I'm not really sure how to go about it (I'm tired and not thinking straight right now, lol).



@Admiral, please use the question forms! c: its easier for others to tell the difference between a question, and a random conversation people have been going on about here. thankyoou c:
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Re: HOMEWORK help center ||HWH||open!

Postby sword » Tue Sep 15, 2015 1:03 pm

Waki - Dang it, sorry. I was in a rush and must have missed that. Thanks for letting me know.
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Re: HOMEWORK help center ||HWH||open!

Postby soymonky » Tue Sep 15, 2015 1:06 pm

omg ;o; this is such a great idea (my parents think this is a stupid game AND NOW I HAVE PROVED THEM WRROONNGG mauhahaha)
(btw how do I add this to my bookmarks ;-;)
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Re: HOMEWORK help center ||HWH||open!

Postby Kylee Hart » Mon Sep 21, 2015 5:14 am

amirite. wrote:omg ;o; this is such a great idea (my parents think this is a stupid game AND NOW I HAVE PROVED THEM WRROONNGG mauhahaha)
(btw how do I add this to my bookmarks ;-;)

Down at the bottom bar (If you scroll all the way down) there's a button that says add to my bookmarks.
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Re: HOMEWORK help center ||HWH||open!

Postby meowool » Tue Sep 22, 2015 5:09 am

Admiral wrote:
Image
TYPE OF QUESTION: Math
YOUR QUESTION: If there's an area that's 10,000 x 10,000 feet on the surface, and 100 feet deep (so, the total volume of that would be calculated in cubic feet), how many tons would that be (volume, not weight)? I'm trying to calculate that because I need to know how much it would cost to mine an area that big for an assignment. I'm not really sure how to go about it (I'm tired and not thinking straight right now, lol).


Pretty simple - volume = depth x width x length Make sure it's all in the same unit and then the answer is 1x10^10 ft^2.

Weight - Tons isn't a measure of volume...?
That depends of what is in that area - with water one square cm weighs one gram* so 1000cm^2 = 1kg, etc...I'm sure there's a way to work it out in imperial - but I never need to use it, and I don't know the US' imperial system anyway. If it's 'earth' then it depends on the density of the rock in that area. (I assume they are using real life geography as that would be a right pain to work out where I live.)

*This is the basis of the metric system...more or less. They are hopefully trying to redefine a kilogram so it's not based on a physical weight because at the moment, all their standard kilograms keep changing weight/mass!
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