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=sin^2(90-27)+sin^227/cos^2(90-73)+cos^273=cos^227+sin^227/sin^273+cos^273now cos^2A+sin^2A=1hence1/1=1

as A+B =90

secB =sec(90-A) = cosecA = 4/3 {because sec(90-x) =cosec(x)}

b is a right answer ok

B is the correct answer

option B is correct answer.

B.-x^2y^2÷50 is the correct answer

LHS= (tanA + cosecB)^2 - (cotB-secA)^2= (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A)= tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step:(sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA= 2tanAcosecB + 2cotBsecA= 2(tanAcosecB + cotBsecA)

RHS= 2tanAcotB(cosecA + secB)= 2tanAcotB((1 / sinA) + (1 / cosB))= 2(tanA / sinA)cotB + 2tanA(cot B / cosB)= 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB))= 2(1 / cosA)cotB + 2 tanA(1 / sinB)= 2secAcotB + 2tanAcosecB= 2(secAcotB + tanAcosecB)

LHS = RHS

Sorry I don't know answer

LET 2^x=3^y=(1/6)^z=k

Therefore,

K^(1/x)=2

K^(1/y)=3

K^(1/z)=(1/6),Multiply these 3 equations to get:

K^[(1/x)+(1/y)+(1/z)]=1

Hence we can conclude that either k=1 ( which is totally false) or power is equal to “0”.

The correct answer is :[(1/x)+(1/y)+(1/z)]=0

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Photo sahi bhej de samajh me Nahi aaraha hai

Number of cycles per second is Frequency

Given:- the volume of the cubical box is 64 cm³

Let the edge of the cubical box be a cm

=>a³=64=>a=4

∴Edge of the cubical box= 4 cm

volume=(side)^364 cm^3=(side)^3(4cm)^3=(side)^3side=4cm

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Volume of cubical box = (Side)³

64 = (Side)³

Side = ∛64

= 4 cm(A) 4 cm

*Unit given in the question is wrong*

Volume=side³Volume=64cm³64=side³side=4cm

Option A is correct

tnxs

(2x+3y+4z)(2x+3y+4z) = 2x*2x+3y*3y+4z*4z+2(2x)(3y)+2(3y)(4z)+2(4z)(2x) = 4x*x+9y*y+16z*z+12xy+24yz+16xz

answer not open

let x=p/qbe the given rational number.

If the prime factorization of q is of the form 2^nx5^m, where n, m are non-negative integers. Thenxhas a decimal expansion which terminates.

Ans :- ◾️Bakhar is an important type of historical document written in Marathi language.

◾️It is the earliest Marathi genre of the medieval period.

◾️There were almost 200 bakhars written between the seventeenth and the nineteenth century.

◾️The most important of them was about Shivaji.

◾️Bakhars are considered as assets that describe the Marathi history and the criticism in that period.

◾️They were written in Marathi prose and enhanced the patriotism of the people.

◾️Earlier, Persian words were used to but later it was written in Sanskrit.

◾️Some common bakhars are Sabhasad bakhar, Mahikavatichi bakhar, Kalmi bakhar, Chitnis bakhar, Peshwayanchi bakhar and Bhausahebayanchi bakhar.

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x+ 5×2=15x+ 10=15x=15-10x=5

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5 person can save 15 litres of waterthen one person can save = 15/5= 3 litrenow 25 persons can save = 25*3= 75 litres of waterPlease like the solution 👍 ✔️

√( 1-cos theeta)^2/√1-cos^2 theeta= √((1-cos theeta/sin theeta)^2= 1-cos theeta/ sin theta

cosectheta - cot thetathanksplease like the solution 👍 ✔️👍

x²-(√3+1)x+√3=0

x²-x-√3x+√3=0

x(x-1)-√3(x-1)=0

(x-1)(x-√3)=0

x=1,√3

If 5 people can save 15 litres of waterThen 25 people can saveas 5*5==15*5litres25people===75litres

Given equation :a = bq+r

As per Euclid division lemma r should be larger or equal to zero and smaller to bThus, 0<=r<b.

For b=3a=3q+r.0<=r<3r=0,1,2

Therefore possible values of r are 0,1,2.

any another method

यदि एक्स माइनस वन p(x)का गुणनखंड हैput x= 1p(1)= 1+1+k= 0k= -2

2)k-3+k= 02k= 3k= 3/2