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heureka

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 #1
avatar+26396 
+5

\sqrt{15/4-2i}

 

z=a+bi|z|=a2+b2Re(z)=az=|z|+Re(z)2±i|z|Re(z)2

 

z=1542ia=Re(z)=154b=2|z|=(154)2+(2)2=(15242)2+4=(15242)2+44242=152+4342=152+434=2894|z|=174z=|z|+Re(z)2±i|z|Re(z)21542i=174+1542±i1741542=17+1542±i171542=3242±i242=3224±i224=328±i28=4±i14=2±i12

 

alle zweiten Wurzeln von 1542i:2+i122i12

 

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24.10.2016
 #2
avatar+26396 
0
24.10.2016
 #2
avatar+26396 
+10

Three equal circles with radius r are drawn as shown,
each with its centre on the circumference of the other two circles.
A, B and C are the centres of the three circles.
Prove that an expression for the area of the shaded region is:

 

A=r22(π3)

 

 

AABC=12rhABCh2ABC+(r2)2=r2h2ABC+r24=r2h2ABC=r2r24h2ABC=34r2hABC=32rAABC=12rhABCAABC=12r32rAABC=r243

 

AArc=πr260360AArc=πr26

 

Ashaded region=3(AArcAABC)+AABCAshaded region=3AArc3AABC+AABCAshaded region=3AArc2AABCAshaded region=3πr262r243Ashaded region=πr22r223Ashaded region=r22(π3)

 

 

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21.10.2016
 #3
avatar+26396 
0

e

Given that (2xi3) divides P(x) find all roots of P(x)

 

 

P(x)=x66x5+172x47x3+212x2+3xP(x)=x(x56x4+172x37x2+212x+3)

 

If (2xi3) divides P(x) then (2x+i3) divides P(x)

 

(2xi3)(2x+i3)=(2x)2(i3)2=2x2(i23)|i2=1=2x2[(1)3]=2x2+3

 

{nl} (x56x4+172x37x2+212x+3):2x2+3=12x33x2+72x+1

 

P(x)=x(2xi3)(2x+i3)(12x33x2+72x+1)

 

test x=2: 12x33x2+72x+1=0

 

(12x33x2+72x+1):x2=12x22x12

 

P(x)=x(2xi3)(2x+i3)(x2)(12x22x12)P(x)=x(2xi3)(2x+i3)(x2)12(x24x1)

 

Roots:

P(x)=0=x(2xi3)(2x+i3)(x2)12(x24x1)1. Rootx1=02. Root2xi3=02x=i3x2=i323. Root2x+i3=02x=i3x3=i324. Rootx2=0x4=25.6. Rootx24x1=0x=4±424(1)2x=4±452x=4±252x=2±5x5=2+5x6=25

 

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21.10.2016
 #3
avatar+26396 
0

3.

man travels 196km by train and returns in a car that travels 21km/h faster.

if total journey took 11hrs, find speeds of train and car

 

Let time train = t1

Let time car = t2

t1+t2=11 ht2=11t1

 

(1)vtraint1=196 kmvtrain=196t1(2)vcar=vtrain+21kmhvcart2=196 km(vtrain+21)t2=196(196t1+21)t2=196|t2=11t1(196t1+21)(11t1)=19619611t1196+211121t1=196|19619611t1196+211119621t1=02156t1196+211119621t1=02156t116121t1=0|t12156t116121t1=0|t12156161t121t21=0|(1)21t21+161t12156=0t1=161±(161)2421(2156)221t1=161±45542t1=161+45542ort1=16145542 no solutiont1=7 hvtrain=196t1vtrain=1967vtrain=28 kmhvcar=vtrain+21vcar=28+21vcar=49 kmh

 

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20.10.2016
 #2
avatar+26396 
0

4.

wire 80cm in length is cut into 2 parts ad each part is bent to form a square.

if sum of areas of squares is 300cm^2, find length of the sides of the square

 

Let first part length = a

Let second part lenght = b

 

Let side of the first square = x

Let sied of the second square = y

 

(1)x2+y2=300(2)a=4xandb=4y(3)a+b=804x+4y=804(x+y)=80|:4x+y=20y=20x(1)x2+y2=300x2+(20x)2=300x2+20240x+x2=3002x240x+202=3002x240x+400=300|3002x240x+400300=02x240x+100=0|:2x220x+50=0x=20±2024502x=20±2002x=20±21002x=20±1022x=10±52x=10+52orx=1052y=20xy=20(10+52)ory=20(1052)y=201052ory=2010+52y=1052ory=10+52

 

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20.10.2016