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A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x² = −1. Because no real number satisfies this equation, i is called an imaginary number.

since, 3+i can be expressed in the form of a+ib where, a is 3 and b is 1, it is a complex no.

Acomplex numberis anumberthat can be expressed in the forma+bi, whereaandbare real numbers, andiis a solution of the equationx2= −1. Because noreal numbersatisfies this equation,iis called animaginary number. For the complex numbera+bi,ais called thereal part, andbis called theimaginary part.

=6/10 x 50=6 x 5 =30

Hence 30 seats were filled out of 50

6/10 x 50 = 6 × 5 = 30

50 seaters bus6/10 seaters are filled = 6/10 × 50 = 6×5=30there are 30 seaters are filled in the bus.so (50-30) 20 people are needed to fill all the chairs.

5/i³ ....(1)

We know, i² = - 2

Multiply and divide by i

5i/i⁴

5i

Plese write the conjugate the complex number on a paper and post it

Conjugate of (1 + i)^2

(1+i)^2 = 1 - 1 + 2i = 2i

Therefore, conjugate is - 2i

Multiply by (1+i) in numerator and denominator both

(1+2i)(1+i)/(1-i)(1+i)

= (1+i+2i -2)/1+1

= (-1 + 3i)/2

As we see,

Real parts is negative while imaginary parts is positive

Hence this will lie in 4th Quadrant

Given, OP = OQ angle(OPC) = angle(OQC) = 90°

In ΔOPC and ΔOQC OP = OQ angle(OPC) = angle(OQC)OC = OCtherfore, By SAS rule,ΔOPC and ΔOQC are congruent. => angle(POC) = angle(QOC)Therfore,OC is the angle bisector of angle(POQ)hence proved.

sinA=3/4cosA=5/4tanA=3/4

sinA=3/4cosA=5/4tanA=3/5

sinA=perpendicular/basep=3, h=4by pythagoras the ram hyponetious =hb=baseh^2=p^2+b^24^2=3^2+b^216=9+b^216-9=b^27=b^2take squre root of both side√7=bcosA= b/h =√7/4tanA= p/b =3/4ans

for cos root 7/4 and tan is 3root by4

sin =P/H then sin=3/4 ,[under root3 square + 4 square]=25 under root then we write normal 5 because square root of 5 is 25 then base is 5. cosA=5/4 and ranA=3/5.

sin A=3/4cosA=5/4tanA=3/4

Sin A=3/4Cos A=5/4tan A =3/4

given sinA = 3/4 = perpendicular/ hypotenuse perpendicular = 3hypotenuse = 4

we have to find base,h^2 = b^2 + p^24^2 = b^2 + 3^216 = b^2 + 916 - 9 = b^27 = b^2

base = root7

sinA = 3/4

cosA = base/ hypotenuse = root7/ 4

tanA = Perpendicular/ base

= 3 / root7

ANSWER sinA=3/4cosA=5/4tanA=3/4

sinA=3/4 लंब/कर्णcosA=5/4tanA=3/4

cosA=✓7/4 tanA=3/✓7

I don't know 🙏 🌹 sorry

cosA)=2.645/4 & tanA=3/2.645

sinA- 3/4cosA- 5/4tanA- 3/4

sinA=3/4cosA=5/4tana=3/4

tan A =3/4........................

sinA=√7/4 tanA=3/√7

sinA=3/4CosA-5/4tanA-3/5

cos A=3/√7 and tan A =3\√7

sinA=3/4casA=5/4TanA=3/5

cosA = √7/4tanA = 3/√7

SinA 3/4CosA 5/4TanA 3/5ha Bhai

cosA=√7/4 &tanA=3/√7 sinA=3/4

sin=3/4 L=3, k=4b=√(3^2+4^2) b=5cosA=A/K=5/4Tan=L/A=3/5

sinA=√7/4

tanA=3/√7

CosA√3/4 TanA3/√7 he is my answers

sinA=3/4 cosA=5/4 tanA=3/5

sin A=3/4cos A=5/4tan A=3/4

Sin A =3/4 Cos A =1/4 Tan A =3

sinA=.3/4 cosA=5/4. tanA=3/4

SinA=3/4CosA=root 7/4TanA=3/root 7

sinA= 3/4...so P-7 , B-3 ,H-4

So cosA=7/4& tanA=3/7

value of cosA=4/5 and tanA=3/4

tan A = 3/√7 and cosA= √7/4

sin A =3/4then base=√7then cos A= √7/4 and tan A= 3/√7

this is wrong question because root value of not applicable on p/q form

ANSWERsinA=3/4cosA=5/4tanA =3/4

cos=root7/4tan=3/root7sin=3/4

cosA= √7/4tanA=3/√7

sinA=3/4CosA=√7/3TanA=3/√7

cosA=√7/4TanA=3/√7sona=3/4

cos = 7/4. Tan= 3/7

sinA=3/4 cosA=5/4 tanA=3/4

CosA=root5/4 tanA=3/root5

Sin A= 3/4 = p/h in pithagorius form ( here p=3 & h= 4)So now b = whole root over of h square - P square = root over of 16- 9 = root 7So cos A = b/h = root7/4 &. Tan A = p/b = 3/root7

sinA=3/4 we know sinA=p/hand h2=p2+b2hereh=4andp=3so,16=9+bb=16_9b=7squrethen cos=root7/4and tanA=3/root7

sinA=3/4CosA=5/4tanA=3/4

we know that sinA =3/4,cosA=√7/4sinA/cosA=3/7

cosA= √7/4 , tanA= 3/√7

cosA=√7/4,and tanA=3/√7

cos A= 5/4tan A= 3/5

sinA=3/4cos=5/4tan=3/4

Diya hai sinA= 3/4 jante Hain formula=Lal beti ka ka

given that sinA=3/4 Cos A =√7/4 and tanA=3/√7

cosA=√7÷4and tanA=3÷√7

cos A=5/4 and tanA =3/5

cosA= root7/4 and tanA=3/root7

sin A=3/4 and cosA=√7/4 and tanA=3/√7

cosA= √7/4

tanA= 3/√7

option ax=a+c-b is correct answer

7x/4(x^2+2/3x+5/6)= 7x^3/4+7/6x^2+ 35/24x

7x/4(x^2+2/3x+5/6)= 7x^3/4+7/6x^2+35/24x

7x/4(x^2+2/3x+5/6)= x^3/3(7/4)+x^2(14/12)+x(35/24)

Asolenoidis a cylindrical coil of wire whose diameter is small compared to its length.

The nature of magnetic field lines produced by a solenoid is similar to that of a bar magnet.the field lines are closed curves, they emerge from the north pole and merge in the south pole outside the magnet and inside the direction is from south to north.inside the solenoid and a bar magnet the field is uniform.the poles in a solenoid can be determined by using a bar magnet or by clock face rule.

A cylindrical coil of copper wire is called solenoid

माध्य = पदो का योग/पदो की संख्या 2 = (1+2+x+3)/4 8 = 6+x x = 2

here sp=sale pricecp=cost pricep%=profit%l%=loss%

GivenLoss percent=15%

Sp=₹4080

Cp=sp*100/100-loss percent

=4080*100/100-15

=4080*100/85

=₹4800

Now,

Cp=₹4800

Gain/profit percent=15%

Sp=cp*(100+gain percent)/100

=4800*(100+15)/100

=4800*115/100

=₹5520

5520 is the right answer

5520₹ is the answer of the following

Rupees 5520 is the answer

Birth rate = 10.7%death rate = 3.2%rate of increase in population = birth rate - death rate = 10.7-3.2 = 7.5%

the population 64000then increase in population in 1st year = (7.5/100)*64000= 4800population at the end of first year = 64000+4800 = 68800

then increase in population in 2nd year = (7.5/100)*68800= 5160population at the end of second year= 68800+5160 = 73960

then increase in population in 3rd year = (7.5/100)*73960= 5547population at the end of third year= 73960+5547 = 79507

So the population at the end of third year is 79507.

Growth rate = 10.7 - 3.2

= 7.5%

= 64000(1+7.5/100)³

= 7765.8

= 7766

population = 64000+7766

= 71766

Given: Birth rate=10.7%Death rate=3.2%Formula used:Rate of increase =birth rate-death rateSolution:rate of increase=10.7-3.2=7.5%increase in population In 1st year=(7.5/100)*6400=4800at the end of 1st year=64000+4800=68800increase in population In 2nd year=(7.5/100)*68800=5160at the end of 2nd year=68800+5160=73960increase in population in 3rd year =(7.5/100)*73960=5547at the end of 3rd year=73960+5547=79507population at the end 3rd year=79507

A because Perimeter is 10*4=40m

and for every 2m posts were placed=20

But 4 posts are repeated so subtract it

So there will be (20-4)= 16 posts.

Like my answer if you find it useful!

b 20because one side of square is 10 then perimeter of square =4×side= 40 and then 2m apart of total pole = 40÷2 = 20 ans.

i) Rome is not in Italy.

ii) 5 + 5 is not equal to 10

iii) 3 is not greater than 4

iv) John is not good in river rafting

thanks

qestion 2 answer is -4

question 1 answer is 25

1 hour =60minutesHence 6 hours=6*60=360 minutes

1 hour = 60 minutes

therefore,

6hours = 6× 60 = 360 minutes

We know 1 dozen = 12 Weight of 12 orange = 1kg760gor 1.7kgWeight of 1 orange = 1.7÷12=0.14Now Weight = 13kgNo.of orange= 13÷0.14=approx. 92 bananas

no.of orange 92 bananas

12 oranges will weigh 760g

speed = distance/time460km/6 hours76.66kmph will be the speed

(x+1/x)²=x²+1/x²+2(x+1/x)²=7+4√3+1/(7+4√3)+2(x+1/x)²=7+4√3+7-4√3+2(x+1/x)²=14+2=16

x+1/x=√16=4

option A is correct

x^4-14x^2y^2-51y^4=x^4-17x^2y^2+3x^2y^2-51y^4=x^2(x^2-17y^2)+3y^2(x^2-17y^2)=(x^2+3y^2)(x^2-17y^2)=(x^2+3y^2)(x+root(17)y)(x-root(17)y)

x³/8+8y³ the correct answer of the given question

5y^2 - 20y - 8z + 2yz= y(5y + 2z) - 4(5y + 2z)= (y-4)(5y + 2z)