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Tutors Answer Your Questions about Quadratic Equations (FREE)
Question 240518: x^2+6x+6=0 solve with quadratic formula
Answer by Alan3354(6082)   (Show Source):
Question 241365: the base of a triangular sail is 20m shorter than it's height.
the area is 150m2. find the base and height of the sail.
Answer by unlockmath(117)  (Show Source):
You can put this solution on YOUR website!Hello,
Let's represent the height of the sail with x and the base will be x-20.
We know the area of a triangle is 1/2(Base)(height), so we can plug in the values given.
1/2(x)(x-20)=150m2. Let's multiply by 2 to remove the fraction.
(x)(x-20)=300m2 Now multiply to give us:
x^2-20x=300m2 Now subtract 300 from both sides.
x^2-20x-300=0 This can be factored into:
(x-30)(x+10)=0 this gives us
x=30 and x=-10
30 Makes sense so the height is 30 meters and the base is 10 meters.
RJ Toftness
www.math-unlock.com
Question 241366: find the vertex, the line of symmetry,
the maxiumum or minimum value of the quadratic
function and graph the function
f(x)=7-x2 and find zeros
Answer by stanbon(26274)  (Show Source):
You can put this solution on YOUR website!find the vertex, the line of symmetry,
the maximum or minimum value of the quadratic
function and graph the function
f(x)=7-x2 and find zeros
You have a quadratic with a = -1, b = 0, c = 7
-----------------
Max or min occurs when x = -b/2a = -0/-2 = 0
f(0) = 7
So the vertex is (0,7)
------------------------------
Axis of symmetry is x=0
-------------------------------
Zeroes: 7-x^2 = 0
x^2 = 7
x = +sqrt(7) ; x = -sqrt(7)
====================================
Graph:
=====================================
Cheers,
Stan H.
Question 241157: "You are told to build a rectangular fence along side a building. You have only 80 feet of fencing. The company wants the maximum amount of area enclosed. Find the dimensions that will lead to the most area"
All I need is the equation and I understand the problem from that point.
Answer by ankor@dixie-net.com(6693)  (Show Source):
You can put this solution on YOUR website!"You are told to build a rectangular fence along side a building.
You have only 80 feet of fencing.
The company wants the maximum amount of area enclosed.
Find the dimensions that will lead to the most area"
:
Along side a building, only 3 sides required, the equation:
L + 2W = 80
L = (80-2W)
:
Area = L * W
A = (80-2W)*W
A = -2W^2 + 80W
Find the axis of symmetry for max area
Question 241069: how to solve this equation using direct variation? 5x-6y=0
Answer by Edwin McCravy(2922) (Show Source):
You can put this solution on YOUR website!how to solve this equation using direct variation? 5x-6y=0
y varies directly with x. So we write
y = kx
We can see that when x=6 and y=5, the equation
5x-6y=0 is true, for 5(6)-6(5)=0
So we substitute these values in
y = kx
5 = k(6)
Divide both sides by 6
Therefore replacing k by  in y=kx we have
y = x
Edwin
Question 241028: Solve the following equation using quadratic formula:
x^2+12+35=0
The solution set is { , }
x=(-12ħ√(〖12〗^2+35))/2
(x+5)(x+7)= 12/2= {5,7}
thanks for you help
Vicki
Answer by rapaljer(3620) (Show Source):
You can put this solution on YOUR website!The quadratic formula is 
In this case, x^2 +12x+35=0, a=1, b=12, and c=35.
 or 
 or 
This problem can be checked by solving by FACTORING:
x^2 +12x+35=0
(x+5)(x+7)=0
x=-5 or x=-7
I have a LOT of resources on my own website on the topic of QUADRATIC FORMULA. To find my website, do a "BING" or a "GOOGLE" search for my last name "Rapalje". Near the top of the search list, you should see "Rapalje Homepage." Click on this, and look near the top of my homepage for the link "Basic, Intermediate and College Algebra: One Step at a Time." Choose "Intermediate Algebra", and look in "Chapter 4" for "Section 4.03, Quadratic Formula." This is my own explanation especially written for students who have trouble understanding math. I wrote this to students, NOT mathematicians, and I think you will find it easier to understand than your own traditional textbooks!! In addition to this page, the entire website is supported by my "MATH IN LIVING COLOR" pages, where the hardest problems are solved IN COLOR.
If you like my website, please recommend it to family and friends who have trouble understanding math!! It's ALL FREE!!!
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
Question 241001: find the root of this equation by factoring:
x+8x=-16
Answer by stanbon(26274) (Show Source):
Question 240939: Polynomials
(y^3)^2
Please help me
Thank you
Answer by edjones(3296)  (Show Source):
Question 240893: Determine the maximum area of a triangle, in square centimeters, if the sum of its base and its height is 10 cm.
Answer by checkley77(7053) (Show Source):
Question 240936: How do you solve the following quadratic equations:
-6(u+5)^2=120
3s^2-1=7s^2
Found 2 solutions by edjones, stanbon:
Answer by edjones(3296) (Show Source):
You can put this solution on YOUR website!-6(u+5)^2=120
(u+5)^2=-20
u^2+10u+25=-20
u^2+10u+45=0
u^2+10u =-45
u^2+10u+25=-45+25 Completing the square
(u+5)^2=-20
u+5=+-sqrt(-20) Take sqrt of each side.
u=-5+-2sqrt(-5)
u=-5+2sqrt(5)i, u=-5-2sqrt(5)i
.
Ed
Answer by stanbon(26274) (Show Source):
You can put this solution on YOUR website!How do you solve the following quadratic equations:
-6(u+5)^2=120
---
(u+5)^2 = -20
u+5 = 2isqrt(5) or -2isqrt(5)
u = -5+2isqrt(5) or u = -5-2isqrt(5)
=========================================
3s^2-1=7s^2
4s^2 = -1
s^2 = -1/4
s = i/2 or s = -i/2
=======================
Cheers,
Stan H.
Question 240910: x^4+10x^2+9=0
solve as quadratic function?
my assumption is
x^2=-1 and x^2=-9
which there is no real solution but i dont think i am right
Answer by Earlsdon(4900) (Show Source):
You can put this solution on YOUR website!Solve:
 Rewrite this as:
 Factor.
 Apply the zero product rule.
 or  so...
If then  so  or 
If then  so  or 
These answers can be written as: (Note:  )
You are correct in stating that there are no REAL solutions, the solutions are complex.
Question 240880: I got this reply back for my question....I can't find an answer unless I am doing something wrong. Thanks for checking into it for me!
Question 240789: Hello,
I need some help to:
1. Solve the following equation using quadratic formula
2x^2-3x-1=0
The solution set is? ________
Thanks!!
Answer by vleith(1971) About Me (Show Source):
Answer by Alan3354(6082) (Show Source):
Question 240861: Use the quadratic equation to solve the equation:
-2.8x= -3.8x^=1.4
Please help me with this, I get stuck everytime when it comes to simplifying and finishing the problem. Thank you so much!
Answer by checkley77(7053) (Show Source):
Question 240782: The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 283 vibrations per second and a wavelength of 12 ft. Fine the wavelength of another tone that has a pitch of 348 vibrations per second.
Answer by Alan3354(6082) (Show Source):
You can put this solution on YOUR website! The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 283 vibrations per second and a wavelength of 12 ft. Fine the wavelength of another tone that has a pitch of 348 vibrations per second.
--------------
p*w = k (some constant)
P*w = 283*12 = 3396
----------
w = 3396/p
w = 3396/348
w = 9.7586 feet
------------------
PS I would call it frequency, not pitch
Question 240809: x^2-4x-12=0
Answer by JimboP1977(76)  (Show Source):
Question 240344: A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral. He has 250 feet of fencing available and wants to use all of it. What is the maximum area of the corral?
There's a picture of it here: http://math.library.wisc.edu/reserves/114/114-nenciu-09spex01.pdf (question #5)
Help? I'm so confused!
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral.
He has 250 feet of fencing available and wants to use all of it.
What is the maximum area of the corral?
:
Four sides will be required, 2 widths and 2 lengths: x and (x+50)
The fence equation:
x + (x+50) + 2W = 250
2x + 50 + 2W = 250
2x + 2W = 250 - 50
2x + 2W = 200
simplify divide by 2
x + W = 100
W = 100-x)
:
Area = x * W; replace W with (100-x)
A = x(100-x)
A(x) = -x^2 + 100x; this should answer part (A)
:
The x value for max area, will be the axis of symmetry: x = -b/(2a)
In this equation a=-1; b=100
x = 
x = 
x = +50; this should answer part(b)
:
Find the dimensions
W = 100 - 50
W = 50' is the width
:
L = x + 50
L = 50 + 50
L = 100'
:
Dimensions for max area: 100 by 50 ft; Part(c)
;
:
Check solutions by finding the total fence length
50 + 100 + 2(50) = 250
Question 240789: Hello,
I need some help to:
1. Solve the following equation using quadratic formula
2x^2-3x-1=0
The solution set is? ________
Thanks!!
Answer by vleith(1977) (Show Source):
Question 240797: I need to a step by step solving this equation:0= x^2+15x-1350 , This is from a word problem using quadratic equation solving cost per person.... I know the answer but I do not know how they got it....help please
Answer by vleith(1977) (Show Source):
Question 240777: Let f(x)=3x^2-8x. find a such that f(a)=-5.
Answer by jsmallt9(593) (Show Source):
You can put this solution on YOUR website!If  , then  . And if f(a) = -5 then  .
Now we just solve for a. Since this is a quadratic equation we will
1) Get one side equal to zero (by adding 5 to each side):
2) Factor (and use the Zero Product Property) or use the Quadratic Formula.
By the Zero Product Property, this or any product can be zero only if one (or more) of the factors is zero. So
 or 
Solving these two equations we get:
 or 
 of (You will get these same answers if you use the Quadratic formula correctly.)
Question 240581: The problem is: (x+5)/3 + 4/x = 5/6 My instructions are to use the quadratic formula to solve. I think I am supposed to multiply by the common denominator, but I am not sure.
Answer by jsmallt9(593) (Show Source):
You can put this solution on YOUR website!You are correct. Probably the easiest way to proceed is to multiply both sides of the equation by the Lowest Common Denominator (LCD) of all the denominators. The LCD of 3, x and 6 is 6x so we'll multiply both sides by 6x:
On the left side we use the Distributive Property to multiply:
The denominator of each fraction cancels with some part of 6x leaving:
Simplifying this we get:
Since this is a quadratic equation we will solve it by
1) Get one side equal to zero
Subtracting 5x from each side we get:
2) Either factor or use the quadratic formula. We can factor out a 2:
As a sum of squares, there is no way to factor the 2nd factor any further. The only solutions to this equation then are the x's that make  equal to zero. But if we think about it, we'll realize that there are no (Real) numbers that will make equal to zero. This is so because  would have to be -12 and there are no Real numbers that you can square and get a negative number like -12.
So your equation has no Real solutions. (If you are supposed to find Complex solutions, just use simplify  .)
Question 240628: x^2-5x+35=0
it has no real solutions but what would be the irrational solution?
Answer by jsmallt9(593) (Show Source):
You can put this solution on YOUR website!This is a job for the quadratic formula. For
the solutions can be found with:
In your equation, 
a = 1
b = -5
c = 35
So the solutions are:
Simplifying inside the square root:
At this point, since there is a negative number in the square root, we know that the solutions will be complex (not irrational). For complex numbers we use "i" which stand for  :
Simplifying the rest we get:
Separating this into two terms and writing this in a+bi form we get:
Question 240648: How do you find the turning point (vertex) of a Quadratic function, not using complete the square?
Answer by rapaljer(3620) (Show Source):
You can put this solution on YOUR website!If the equation is y=ax^2 +bx +c, then the vertex is always at the value where x=-b/(2a). After you find x, substitute this back into the original equation (or let the calcluator do it for you!!) to find y.
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
Question 240509: -9t + 2 =56
Answer by College Student(210)  (Show Source):
You can put this solution on YOUR website!In this equation, we need to find the value of t.
First, we will need to move 2 to the other side by deducting it form both sides of the eqn, then divide by -9 on both sides of the eqn to find the value of t.
.
.
Done! :)
Question 240469: a farmer with 4000 meters of fencing wants to encloce a rectangular plot that borders on a river.
*if the farmer does not fence the side along the river, what is the largest area that can be enclosed?
*provide the area function used and graph it. Identify the maximum and interpret what it means.
Answer by checkley77(7053) (Show Source):
You can put this solution on YOUR website!THIS IS A SPECIAL CASE WHERE THE LENGTH IS TWICE THE WIDTH & THERE IS ONLY 1 LENGTH BUT 2 WIDTHS.
L=2W
2W+L=4,000
2W+2W=4,000
4W=4,000
W=4,000/4
W=1,000 ANS. FOR THE EACH OF THE WIDTHS.
L=2*1,000=2,000 ANS. FOR THE 1 LENGTH.
PROOF:
2*1,000+2,000=4,000
2,000+2,000=4,000
4,000=4,000
PROOF:
1,000*2,000=2,000,000 TOTAL AREA.
TEST:
ADD 2 TO THE LENGTH & SUBTRACT 1 FROM EACH OF THE SIDES.
2002*999=1.999,998
ADD 1 TO EACH OF THE WIDTHS & SUBTRACT 2 FROM THE LENGTH.
1998+1001=1,999,998
Question 240473: A rectangular garden is 20 ft longer than it is wide. Its area is 2925 . What are its dimensions?
Found 2 solutions by checkley77, solver91311:
Answer by checkley77(7053) (Show Source):
You can put this solution on YOUR website!W=L+20
LW=2925
L(L+20)=2925
L^2+20L-2925=0
(L-45)(L+65)=0
L-45=0
L=45 FT. IS THE LENGTH.
W=45+20=65 FT. IS THE WIDTH.
PROOF:
45*65=2925
2925=2925
Answer by solver91311(5072)  (Show Source):
Question 240464: do you now gow to write an equation of a parabola with a max value of 4 when x is 1 and one x intercept at 2... help with this specific and ones similar to it please
Answer by stanbon(26274) (Show Source):
You can put this solution on YOUR website!do you now how to write an equation of a parabola with a max value of 4 when x is 1 and one x intercept at 2... help with this specific and ones similar to it please
----------------------------
Vertex: (1,4)
x-intercept: (2,0)
---------------------------
Form:
(x-1)^2 = 4p(y-4)
Solve for "p":
(2-1)^2 = 4p(0-4)
-16p = 1
p = -1/16
=====================
Equation:
(x-1)^2 = 4(-1/16)(y-4)
x^2-2x+1 = (-1/4)(y-4)
(-1/4)y + 1 = x^2-2x+1
(-1/4)y = x^2-2x
y = -4x^2+8x
---------------------------
=======================================================
Cheers,
Stan H.
Question 240471: A rectangular garden is 20 ft longer than it is wide. Its area is 2925 . What are its dimensions?
Answer by solver91311(5072) (Show Source):
Question 240438: What is the values of x in the equation y=x(5x+1)
Answer by JimboP1977(76) (Show Source):
Question 240431: if a rectangle has an area of 117cm2 and a perimeter of 44cm what is it's length and breadth
Found 2 solutions by checkley77, stanbon:
Answer by checkley77(7053) (Show Source):
You can put this solution on YOUR website!LW=117
L=117/W
2L+2W=44
2(117/W)+2W=44
234/W+2W=44
(234+2W^2/W=44
(234+2W^2)/W=44
2W^2+234=44W
2W^2-44W+234=0
2(W^2-22W+117)=0
2(W-13)(W-9)=0
W-13=0
W=13 ANS.
L*13=117
L=117/13
L=9 ANS
2*9+2*13=44
18+26=44
44=44
PROOF:
Answer by stanbon(26274) (Show Source):
You can put this solution on YOUR website!if a rectangle has an area of 117cm2 and a perimeter of 44cm what is it's length and breadth
------------------------
width = x
length = y
-----------------
Area = xy = 117 cm^2
Perim = 2(x+y) = 44 cm
----------------------------
x+y = 22
---------------------
Solve for y:
y = 22-x
-------------------
Substitute into xy = 117 and solve for "x":
x(22-x) = 117
-x^2 + 22x - 117 = 0
x^2 - 22x + 117 = 0
--------------------
Quadratic formula:
x = [22 +- sqrt(22^2 - 4*117)]/2
x = [22 +- sqrt(16)]/2
Positive solution:
x = [11 + 2] = 13 (width)
-------------------------
Since x+y = 22, y = 22-13 = 9 (length)
============================================
Cheers,
Stan H.
Question 240299: x to the 2 power- 3 - 3/4 = 0
a) solve by completing the square
b) solve using the quadratic formula
(x to the 2 power minus x minus three-fourths equal 0.
Answer by stanbon(26274) (Show Source):
You can put this solution on YOUR website!x to the 2 power- 3x - 3/4 = 0
---
a) solve by completing the square
x^2 - 3x - 3/4 = 0
x^2 - 3x + ? = (3/4)+?
x^2 - 3x + (3/2)^2 = (3/4) + (3/2)^2
(x-(3/2))^2 = (3/4) + (9/4)
(x-(3/2))^2 = 12/4=3
x-3/2 = sqrt(3) or x-3/2 = -sqrt(3)
x = (3/2)+sqrt(3) or x = (3/2)-sqrt(3)
---------------------------------------------------------
b) solve using the quadratic formula
a = 1; b = -3; c = -3/4
---
x = [-b +- sqrt(b^2-4ac)]/(2a)
x = [3 +- sqrt(9 - 4*(-3/4)]/2
x = [3 +- sqrt(12)]/2
x = (3/2) + (2/2)sqrt(3) or x = (3/2)-(2/2)sqrt(3)
x = (3/2)+ sqrt(3) or x = (3/2)-sqrt(3)
============================================
Cheers,
Stan H.
Question 240240: i need the formula for how to find the equation of a curved line graph eg y=_xsquared +_x + _
Answer by Theo(675)  (Show Source):
You can put this solution on YOUR website!If you mean a quadratic equation, the formula is:
x = 
standard form of the quadratic equation is:
where:
a is the coefficient of the term.
b is the coefficient of the x term.
c is the constant.
You solve for x.
The y value is 0 because you set the equation equal to 0.
Your roots are:
(x,y) = (x,0) where you replace x with the values you calculated for x.
The roots can be real or imaginary.
If the roots are real, the graph of the equation crosses the x-axis at y = 0.
If the roots are not real (contain an imaginary part which is a negative square root) then the graph of the equation will not cross the x-axis.
The quadratic formula works on all quadratic equations whether the roots are real or imaginary.
Question 239651: What happens when b=0 in the quadratic x^2+bx+c=0?
Answer by College Student(210) (Show Source):
Question 240199: Evaluate each given value of the function f(x) if f(x) =6-2x
Find f(0)
Find f(-2)
Answer by solver91311(5072) (Show Source):
Question 239701: use the discriminate to find the number and type of solutions to the following quadratic equation x^2+2x+6=0 i got the answer of -20 i just don't know if it's 2 real solutions, 2 imaginary solutions or 1 real solution.
Answer by philline_palana(12)  (Show Source):
Question 240082: 3x^2=2x+2
Answer by philline_palana(12) (Show Source):
Question 240118: 3x^2-14x+8=0
Answer by philline_palana(12) (Show Source):
Question 240171: determine how long the cannon ball was in the air.
using h=-5t(squared)+ 40t+ 3
Answer by philline_palana(12) (Show Source):
Question 239988: Doing a quadratic equation problem, i cam across a problem. the equation i got was x^2-6c-3=0. When plugging this problem into the equation, x=-b+/-(sqrt(b^2+4ac))/2a, I did not know if you could plug the b value (-3) into the place of -b or if you had to make it a positive b.
Found 2 solutions by solver91311, JimboP1977:
Answer by solver91311(5072) (Show Source):
Answer by JimboP1977(76) (Show Source):
You can put this solution on YOUR website!Well don't forget that the general form is 
So the value of b in your equation is -6 so (-1)(-6) = 6
so b = 6 in this instance.
Does that help?
Question 239994: 3x^2-5x+4=0
Answer by JimboP1977(76) (Show Source):
Question 239793: hi sir. how we can make equation from word problems
Answer by College Student(210) (Show Source):
You can put this solution on YOUR website!Word problems tend to be tricky and it takes a while to get used to them, but here are a few hints.
.
If the problem tells you there are several consecutive integers, write the following on your notebook:
x = integer
x+1 = consecutive integer
x+2 = next consecutive integer ... continue for as many consecutive #s you're given in the problem.
.
Be careful, though. Some problems want you to look for several consecutive EVEN integers. In which case, you'd write on your notebook:
x = integer
x+2 = consecutive even integer
x+4 = next consecutive even integer ... again, continue for as manyas you need.
.
The problems start telling you how you add, substract, multiply, or divide (+-*/) two of the integers.
.
Then it probably also gives you another (+-*/) wording.
.
The words "equal" and "is" mean the symbol =
.
The word "least integer" would refer to the value of x. Whereas the "greatest" or "largest" integer would be the highest value of x you noted on your notebook (ex. x+4).
.
When you (+ - * or /) your integers, remember to write them as stated above. For instance, if the problem tells you "the third integer is twenty seven more than 3 times the least integer"... you would write: (x+4)+27=3(x).
.
Once you get the equation, work out the algebra and solve for x. This gives you the first integer. Then it would be very easy to figure out the rest of the integers.
.
Always check your answers by pluggin the value of x in the original equation to ensure both sides of the equation are equal to each other. If they are not, then you did something wrong.
.
Good luck!
Question 239736: How would I find the real-number solutions of the equation: x^3-3x=0
Answer by Theo(675) (Show Source):
You can put this solution on YOUR website!x^3 - 3x = 0
you can factor out the x to get:
x * (x^2-3) = 0
Either x is 0 or (x^2-3) is 0
x = 0 is one of the solutions.
x^2 - 3 = 0
add 3 to each side to get:
x^2 = 3
take square root of each side to get:
x = +/- sqrt(3)
those are your solutions.
x = 0
x = sqrt(3)
x = -sqrt(3)
a graph of your equation looks like this:
+/- sqrt(3) is roughly +/- 1.7632 as shown on the graph.
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395
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