took out old questions
edited on feb 19 09
could someone please factor this for me:
2x^2-7x+3
Thanks.
ill rep.
AND HERE ARE MORE:
http://answers.yahoo.com/question/in...9202742AAwxbLm
took out old questions
edited on feb 19 09
could someone please factor this for me:
2x^2-7x+3
Thanks.
ill rep.
AND HERE ARE MORE:
http://answers.yahoo.com/question/in...9202742AAwxbLm






SCAR Code:s<2 = s<4 //opposites are equal
x - 21 = 3x -5
x = 16
s<2 = 4(16) - 21
= 43 //size of angles 2 and 4 is 43 degrees
s<3 = s<1
360 = s<1 + s<3 + s<2 + s<4 //all angles must equal 360 degrees
= 2*s<1 + 2*s<2 //since opposites are equal
= 2s<1 + 2(43)
= 2s<1 + 86
2s<1 = 360 - 86
= 274
s<1 = 274/2
= 137 degrees //ANSWER
I'm pretty sure its correct. Narcle's working is also correct except 4*16 - 21 = 43 not 42![]()
m<2 = M<4
4x-21=3x-5
x-21=-5
x=16
4*16-21=42
180-42=138
m<1=138
I think, so many people on this thread atm lol
(Scripts outdated until I update for new SRL changes)
AK Smelter & Crafter [SRL-Stats] - Fast Fighter [TUT] [SRL-Stats]
If you PM me with a stupid question or one listed in FAQ I will NOT respond. -Narcle
Summer = me busy, won't be around much.
(Scripts outdated until I update for new SRL changes)
AK Smelter & Crafter [SRL-Stats] - Fast Fighter [TUT] [SRL-Stats]
If you PM me with a stupid question or one listed in FAQ I will NOT respond. -Narcle
Summer = me busy, won't be around much.
K I Posted Next Question Please Help On That Too !
REP++ TO ZEP THANKS! for other formula tho narcles shorterand easier thanks again
Ok this is how you do the second question.
Since all sides are equal to each other you seperate the x and y:
First step we find out what x is:
2x-6 = 5x
2x-5x-6=0
2x-5x=6
-3x=6
x=-2 (Divide both sides by -3)
Second Step we find out what y is:
5y = 7y+7
5y-7y=7
-2y=7
y=-7/2 (Divide both sides by -2)
So the answer would be
x= -2, y= -7/2
*If you want to find the length of each side
just substitute the x with -2 and the y with -7/2*
Cheers!![]()






SCAR Code:5x = 7y + 7 2x - 6 = 5y
x = (7y + 7)/5
Substitute into the other equation:
2(7y + 7)/5 - 6 = 5y
(14y + 14)/5 - 6 = 5y
(14y + 14)/5 = 5y + 6
14y + 14 = 25y + 30
11y = -16
y = -16/11
x = (7y + 7)/5
= (7(-16/11) + 7)/5
= -7/11
*Hint: use a graphics calculator and solve use the simultaneous solver.
EDIT:
Are they all equal?






No it means its a right angle. But a rectangle has right angles in every corner yet it doesn't have to have equal sides.
thanks foolish
you deserve srl junior ill rep you thansk alot
i think you did it wrong
its
2x-6=5y
and
7y+7=5x
and i dont know how to solve that...
i still didnt get it seeing zephs explanation :S
I was getting -7/11 for X as well.
Does it fit into it?
(Scripts outdated until I update for new SRL changes)
AK Smelter & Crafter [SRL-Stats] - Fast Fighter [TUT] [SRL-Stats]
If you PM me with a stupid question or one listed in FAQ I will NOT respond. -Narcle
Summer = me busy, won't be around much.
[QUOTE=ZephyrsFury;354539]SCAR Code:5x = 7y + 7 2x - 6 = 5y
x = (7y + 7)/5
Substitute into the other equation:
2(7y + 7)/5 - 6 = 5y
(14y + 14)/5 - 6 = 5y
(14y + 14)/5 = 5y + 6
14y + 14 = 25y + 30
11y = -16
y = -16/11
x = (7y + 7)/5
= (7(-16/11) + 7)/5
= -7/11
*Hint: use a graphics calculator and solve use the simultaneous solver.
QUOTE]
Just look at how Zephyr did it. His answers are correct. P1nky, you can look at mine only if there is going to be square shaped questions on the test. Good Luck tomorrow!
Lol I'm in Grade 7 but im justing learning this stuff.It suczors.If you need more help just PM me I'll be glad to help
Yeah, so on the bottom one, if you still dont understand Zephyrs's:
bottom side = top side
left side = right side
so then:
2x-6 = 5y
5x = 7y+7
So solve for one variable on both of the equations
(top) 2x = 5y-6 ------> x = (5y-6)/2
(bottom) x= (7y+7)/5
since you know that x = x because they are the same variable.....
(7y+7)/5 = (5y-6)/2
and now solve for y (in this case I would cross multiply)
5*(5y-6) = 2*(7y+7)
distribute the coefficients.....
25y-35 = 14y + 14 //////////OMG!!!!!!!! 5*6 doesn't equal 35!!!!![]()
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I hate myself.......... so I'll restart with the right equation..... at the bottom, so you can see my mistakes.....
25y-35 = 14y+14
solve for y
25y = 14y+14+35
25y-14y = 14+35
11y = 49
therefore:
(((((((((((((((((((((((((((((((y = 49/11)))))))))))))))))))))))))))))))))))))
now that you have y, stick it one of your other equations and get x
7y+7 = 5x
7(49/11)+7 = 5x
343/11 + 77/11 = 5x 77/11 = 7
420/11 = 5x divide by 5, and you're golden
420/55 = x (might wanna simplify)
(((((((((((((((((((((((((((((((x = 84/11)))))))))))))))))))))))
hope I'm right.........wouldn't look good if a kid in calc II failed geometry.....
EDIT!!!!
25y-30 = 14y+14
solve for y
25y = 14y+14+30
25y-14y = 14+30
11y = 44
therefore:
(((((((((((((((((((((((((((((((y = 4)))))))))))))))))))))))))))))))))))))
now that you have y, stick it one of your other equations and get x
7y+7 = 5x
7(4)+7 = 5x
28 + 7 = 5x
35 = 5x divide by 5, and you're golden
7 = x
(((((((((((((((((((((((((((((((x = 7)))))))))))))))))))))))))))
sorry for the mistake...stupid me...stupid multiplication!![]()
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OK GUYS i edited and added new questions i need help with
THANKS ILL REP I NEED IT BY 2NITE lol test 2mooro
thanks again
cmon anyone?
1)
The first one is a right triangle. It is because when you plug the two smaller numbers into the Pythagorean Theorum(A^2 * B^2 = C^2) It gives you the third number
Now the second part is acute because the two smaller numbers squared and added together is 181. Then, 13 squared is 169. Since the square root of 181 is bigger than the square root of 169, the triangle is acute, because to be right, that third large side would have to be the square root of 181 long.
2) I am 90% sure the area is 90.
Ok. If you took the places where the lines come together on the side and drew from there to the center point, all the way around, they would all be equilateral triangles(equal side lengths). Knowing that, all we need is the height if that line drew in to the center on your picture, and we know the area of that 1 triangle. Multiply times 6 to get the whole thing.
-Illustration:
-------/|\
-----/--|-\
-A--/--|--\
---/----|---\ B
--/---D|----\
-/------|-----\
-___________
-------C
(Had to put in the hashes so it would post right)
Lines A, B, and C are all the same length, 6. So, we need to find the length of line D. Ok, so using the pythagorean theorum, A(Hypotenuse) squared is 36, minus half of C which is 3, so 9 when you square. 36 - 9 = 25. Square root is 5. That is the length of line D. So, to find the area of the triangle, take half of C (3) times the height D (5). You get 15 as the area of the triangle.
Now, there are 6 triangles in your polygon, so multiply times 6, and you get 90!
Hope I made sense with all that. Feel free to ask questions.
- 9th Grader
REP++ thanks but the second 1 was kind of confuzing :S
ADDED ANOTHER QUESTION
Ok, can you give me any more of a specific are that you can't get?
You are basicly finding that length of the line drawn up to the center, in order to find the area of 1 section (triangle) with the pythagorean theorum.
Then you multiply by the number of sections (triangles) you have in your polygon(also the number of sides) to find the total area.
I gotta get to bed, but i'll be back tomorrow, I promise.
alrite goodnight hopefully someelse can help for some other questions
thanks rockman i think i get the second not sure though![]()
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