Membres: admin

Mathefreak mal pi:
Aber wenn du X als Zahl definierst, kann es unmöglich sein ohne gegennenner eine Zahl zu teilen ohne es mit einer Variable anzuschreiben

admin4 mars 2012

+3146

wahaby:
5 kg of apples cost £7
2kg of apples and 3kg of bananas cost £5.65
work out the cost of 1 kg bananas

a: cost of 1kg of apples
b: cost of 1kg of bananas

searched: b = work out the cost of 1 kg bananas

equation1: 5*a = 7
equation2: 2*a + 3*b = 5.65

[input]solve( 5*a = 7, 2*a + 3*b = 5.65 )[/input]
{a=1.4, b=0.95}

wahaby:
work out the cost of 1 kg bananas

b= £0.95

test: 5*a = 5*1.4 = 7
test: 2*a+3*b = 2*1.4+3*0.95 = 5.65

solve manually:
5*a = 7
a = 7/5

2*a + 3*b = 5.65
2*(7/5) + 3*b = 5.65
14/5 + 3*b = 5.65
14/5 + 3*b -14/5 = 5.65 - 14/5
3b = 2.85
b = 2.85 / 3 = 0.95

admin4 mars 2012

+3146

[input]2 3/4[/input]

admin4 mars 2012

+3146

you are really using "x" as variable and for "times"
(3x3 - 4x2 + x - 2) - (x3 - 5x2 + 3x + 2)

(3*3 - 4*2 + x - 2) - (x*3 - 5*2 + 3*x + 2)
(9 - 8 + x - 2) - (x*3 - 10 + 3*x + 2)
(x - 1) - (3*x - 8 + 3*x)
x - 1 - 3*x + 8 - 3*x
x - 6*x + 7
7-5x

admin4 mars 2012

+3146

[input]4(x - 3) = 12[/input]

4(m - 3) = 12
4m-3*4=12
4m-3*4+3*4=12+3*4
4m=12+3*4=12+12=24
4m/4 = 24/4
m = 6

admin4 mars 2012

+3146

TIARRA:
There are 6 cards, 2 with stars on them and 4 with circles on them. Two cards are chosen at random.

All of the following probabilities represent dependent events in this scenario except:

A) picking a star, keeping it, and then picking another star.
B) picking a star, replacing it, and then picking a circle.
C) picking a circle, not replacing it, and then picking a star.
D) picking a circle, keeping it, and then picking another circle.

PLEASE CAN U HLP ME IN THIS ONE REALLY FAST??

admin4 mars 2012

+3146

Gary89:
Welche Zahl kann man durch 39 teilen ?

jede.

admin3 mars 2012

+3146

usman khan:
show that the function y=x^3-3x^2+5 has a maxima at x=0 and minima at x=2

[input]plot( x^3-3x^2+5, x=-1..3)[/input]

[input]diff( x^3-3x^2+5, x)[/input]
[input]3*x^2-(6*x)=0[/input]
so f'( 0 )=0 and f'( 2 )

[input]diff( diff( x^3-3x^2+5, x))[/input]
[input]6*x-6=0[/input]
and f''( 0 ) <> 0, and f''( 2 )<>0

admin3 mars 2012

+3146

Fred:
-5[ 4x-(x+2)]

-5[ 4x-(x+2)]
-5[ 4x-x-2]
-5[ 3x-2]
-5* 3x+10
-15x+10

admin3 mars 2012

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