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Algebraic Rating of Warlight: 5/14/2017 11:57:18  Ashmage
Level 48
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Please represent your rating of Warlight in an algebraic expression in 1 to 10000. 10000 is the best.
Algebraic Rating of Warlight: 5/14/2017 12:18:32  Japanball
Level 56
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6921
Algebraic Rating of Warlight: 5/14/2017 14:04:16  DanWL Level 62
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9x - 15y = 1050
2x + 2y = 500

z = xy

Work out z (in number form).

Edited 5/14/2017 14:05:23
Algebraic Rating of Warlight: 5/14/2017 14:20:27  OnlyThePie
Level 54
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z = 10000
Algebraic Rating of Warlight: 5/14/2017 16:24:34  Japanball
Level 56
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9x-15y=1050
2x+2y=500

18x-30y=2100
18x+2y=4500

a+(b+32)=c+2400
a-b=c
2b+32=2400
2(30y)+32=2400
60y+32=2400
60y=2368
I went wrong somewhere
Algebraic Rating of Warlight: 5/14/2017 18:07:31  OnlyThePie
Level 54
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2x=500-2y
x=250-y
9(250-y)-15y=1050
2250-9y-15y=1050
-24y=-1200
y=50
2x=2(50)=500
2x=400
x=200

z=(200)(50)
z=10000
Algebraic Rating of Warlight: 5/14/2017 19:03:51  apollong3
Level 53
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18x - 30y = 2100 (1)
-18x - 18y = -4500 (2)

(1)+(2) => -48y = -2400 => y= 50

(2) => 2x + 100 = 500 => 2x= 400 => x=200

z = 200*50 => z= 10000

When you multiply (2x + 2y = 500) with 9, remember to multiply everything with 9
Algebraic Rating of Warlight: 5/14/2017 19:32:04  apollong3
Level 53
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If:

lnx -ln3.5 -ln2 +log(10^1/128) = 1/2^x (x > 0)

y=1000x

find y to reveal my rating

(note that e.g. x^? means "x" to the power of "?")

Edited 5/14/2017 19:32:37
Algebraic Rating of Warlight: 5/14/2017 23:10:44  DanWL Level 62
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a = (b+c)/c - (b+c) - 3c
b = 2c
(c/5) * √25 = 0.75 + 1/4
d = ((2b+c) - a) * 10^3
e = d/(4c) + d

Edit: typo - before e = d/(2c) + d

Edited 5/15/2017 09:47:55
Algebraic Rating of Warlight: 5/15/2017 02:06:53  Benjamin628 Level 59
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x * [e^(pi*i)] = -10000

Edited 5/15/2017 02:07:21
Algebraic Rating of Warlight: 5/15/2017 06:16:47  apollong3
Level 53
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so:
(1)=> lnx - (ln3.5 +ln2) -1/2^x = -log(10^1/128)

however:
ln3.5 +ln2 = ln(3.5*2) = ln7

and:
log(10^1/128) = 1/128 = (128 = 2^7) = 1/2^7

so:

lnx -1/2^x = ln7 -1/2^7 (3)

I now define a function f(x) = lnx -1/2^x

I will skip the proof, but just take it that f(x) is strictly increasing(that means that for every x,y that belong to (0, +infinity), with x < y , f(x) < f(y). This also means that: x = y <=> f(x) = f(y) )

(3) => f(x) = f(7) => x = 7

that means y = 7*1000 => y = 7000

EDIT:
http://prnt.sc/f7xz7t

eh, that's pretty much accurate

Edited 5/15/2017 06:47:31
Algebraic Rating of Warlight: 5/15/2017 07:18:08  apollong3
Level 53
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That is indeed what I meant.
Algebraic Rating of Warlight: 5/15/2017 09:48:50  DanWL Level 62
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c = 1
b = 2
a = -3
d = 8000
e = 10000

Let me show how:
(c/5) * √25 = 0.75 + 1/4
(c/5) * 5 = 1
the divide and multiply cancel out...
c = 1

b = 2c
b = 2*1 = 2

a = (b+c)/c - (b+c) - 3c
a = (2+1)/1 - (2+1) - 3*1
a = 3/1 - 3 - 3
a = 3 - 3 - 3
a = 0 - 3
a = -3

d = ((2b+c) - a) * 10^3
d = ((2*2+1) - -3) * 1000
two negatives make a positive, so - -3 = +3
d = ((4+1) + 3) * 1000
d = (5+3) * 1000
d = 8 * 1000
d = 8000

e = d/(4c) + d
e = 8000/(4*1) + 8000
e = 8000/4 + 8000
e = 2000 + 8000
e = 10000
Algebraic Rating of Warlight: 5/15/2017 10:04:36  Krzysztof Level 66
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z = DanWL,
knowing that D = 44, anWL is equal to fat black man eating pizza with onion, strawberry and pudding, John F. Kennedy was assassinated by mad cow and Vladimir Putin is twin sister of Julius Ceasar - solve X
Algebraic Rating of Warlight: 5/15/2017 11:15:36  DanWL Level 62
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^
X >= 1 and X <= 10000
because this thread says that the number can only be between 1 and 10000, x is somewhere in between.
That's all you wanted me to solve

Solve (Krzysztof)/180540 to zero decimal places. The answer should be between or equal to 9000 and 10000.

K = f/25
r = tK (to two decimal places)
z = 4K ± 10
y = √42 (to two decimal places)
s = zy (to two decimal places)
t = oy - (z/2) (to two decimal places)
o = 2y (to two decimal places) ± 3
f = 5*-1

Edited 5/15/2017 13:32:33
Algebraic Rating of Warlight: 5/15/2017 12:02:20  Ashmage
Level 48
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DanWL that's clever.
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