All Answers 358179 Answers

+3146

0

[input]16 - 6 / 7 - 7[/input]

adminOct 1, 2012

+3146

0

[input]solve( x/5-g=a, x)[/input]

adminOct 1, 2012

+3146

0

[input](x-3)^4+ 30=0[/input]
no real solutions.

[input](x-3)^4-30= 0[/input]
two real solutions.

(x-3)^4-30= 0
(x-3)^4-30+30= 0+30
(x-3)^4= 30
sqrt4((x-3)^4)= sqrt(30)
(x-3) = sqrt4(30) and (x-3) = -sqrt(30)
x = sqrt4(30)+3 and x = -sqrt(30) + 3

adminOct 1, 2012

+3146

0

[input]plot( -2x^2+ 12, 3x, x=-5..5 )[/input]

[input]-2x^2+ 12 = 3x[/input]

x<=-(1/4*sqrt(105))-3/4
x>=1/4*sqrt(105)-3/4

adminOct 1, 2012

+3146

0

[input]x*17%=132[/input]
[input]1%*13200/17[/input]

adminOct 1, 2012

0

weiß ich auch nicht aber im internet findet man so gut wie alles.

GuestOct 1, 2012

0

Trick question? lol 5

GuestOct 1, 2012

Sep 30, 2012

+14

0

If you can tell me exactly what you want to factorization may be able to help

HellasSep 30, 2012

+14

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what you think you can do is (4ψ^2-3ψ-5)/2ψ^2+2ψ-2)
=[2ψ(2ψ-1)-ψ-5]/[2ψ(ψ+1)-2]

HellasSep 30, 2012

+14

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8-4x<3-5x => 5χ-4χ<3-8 => χ<-5

HellasSep 30, 2012

+14

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The correct answer is χ^3-χ^2χ+1 => χ^2*(χ-1)+χ-1 => (χ^2+1)*(χ-1)

HellasSep 30, 2012

+14

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if you mean i^17
17/4=16 with remainder 1 so i^17=i^1=i

HellasSep 30, 2012

0

its me again, i mean i want to know how to simplify this:

V
___

b/3

GuestSep 30, 2012

0

its me again, i mean i want to know how to simplify this:

V
___

b/3

GuestSep 30, 2012

0

X1=0.16*X1+0.15*X2+195 => XI*(1-0.16)= (0.15X2+195) =>
X1 = (0.15X2+195)/(1-0.16) also durch den X1 ersetzt (0.15X2+195)/(1-0.16)
in Relation х2=0.27*х1+0.35*х2+260 Immer, wenn ich
χ2=0,27*(0,15*χ2+195)/(1-0,16) +0,35*χ2+260
=> χ2= (0.0405*χ2+52.65)/0.84 +0,35*χ2+260
=>0,84*Χ2= 0,0405*Χ2+52,65 +0.294*Χ2 +218.4
=> 0.5055*Χ2 = 271.05 => Χ2=536.20178041543027 Abgerundete WIRD Χ2=536

Χ1=0,16*Χ1+0,15*Χ2+195 => Χ1-0,16*Χ1=0,15*536.20178041543027+195
=> Χ1*0,15=275.4302670623145405 => Χ1=1836.20178041543027

SONST abgerundeten IS 0,15*Χ1=0,15*536+195 =>0,15*Χ1=275,4 =>Χ1=1836

GuestSep 30, 2012

0

Generally the function f (x) = sinx defined for each x ∈ R
ie the function sine has the domain R
the whole of the function y = -2 sin pi x is [-2, 2], which means that it has minimum and maximum 2 -2
This function also is redundant ie f (-x) = - sin pi x
Because it has pi x

generally if a real number x that verifies the equation sin x = y with -1 <y <1 then apply
sin x = y <=> x = 2 kpi + x or x = 2k pi + (pi-x)

Example To solve the equation sin x = 1/2 is first a corner on the interval [a, 2 pi] which has sine 1/2
We know that 1/2 = sin pi / 6 so we
sin x = 1/2 <=> x = 2 x pi + pi / 6 or
x = 2 x pi + (pi-pi / 6) = 2 k pi +5 pi / 6 with k belongs to Z

The trgonometria requires much reading but rather to solve exercises

GuestSep 30, 2012

0

No, but thanks anyway. That's not the answer the book gave. The book gave as the answer (x+1) (x-1)^ But we have to show how we got the answer.

GuestSep 30, 2012

Sep 29, 2012

0

Ok this is the exercise y= -2 sin pi x maybe this way you can understand it

GuestSep 29, 2012

0

tcc1:
how do you work the following: y=-2sinpix

If you can tell me what it is sinpi then maybe i can solve this THANKS

GuestSep 29, 2012

0

3x/5-x=x/10-5/2 =>3X/5-X-X/10=-2/5 =>10*3χ/5-10χ-10χ/10=-2*10/5 =>6χ-χ-10χ=-25 =>-5χ=-25 => χ=5

Least common multiple=10

GuestSep 29, 2012

0

2X+1>5 => 2X>4 => X>2

GuestSep 29, 2012

0

should x> 0
x*lnx=x-2 => lnx=(2-x)/x => lnx=ln(2-x)/x => x= (2-x)/x => x^2+x-2=0 ,Δ=1^2-4*(-2)*1=1+8=9
x=(-1*-3)/2 and x=(-1+3)/2 so x=1 and x=-2

GuestSep 29, 2012

0

GuestSep 29, 2012

+10

0

k42Sep 29, 2012

0

Amplify 1st fraction with 8 and 2nd with 3.
-2/3 + -7/8 = -16-21/24 = -37/24

GuestSep 29, 2012

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