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4 is a right answer ok.

If three points (x1, y1), (x2, y2) and (x3, y3) are collinear then [x1(y2- y3) + x2( y3- y1)+ x3(y1- y2)] = 0.A( -1, -4),B(b,c) and C(5, -1) arecollinear then [x1(y2- y3) + x2( y3- y1)+ x3(y1- y2)] = 0.⇒ [ -1( c + 1) +b( -1 + 4) + 5(-4 - c)] = 0.⇒[ - c - 1 -b + 4b - 20 - 5c] = 0.⇒[3b - 6c] = 21⇒b - 2c = 7 -------(1)Given2b + c = 4 -------(2)solving (1) and (2) we getb = 3 and c = -2.

according to theorem these two angles will be equalso 5x+11= 2x+353x= 24x= 8°value of b = (2*8+35)16+3551°

i dont know answer sorry and thanku

x + y = 4 .....(1)

2x - 5y = 1 ......(2)

5 × (1) + (2)

7x = 21

x = 3

So, y = 4 - 3 = 2

(i) Line segment(ii) Line(iii) Plane(iv) Ray

let the 4th proportion = x=3:5::27:x= 3/5 = 27/x= 27/x = 3/5= 1/x = 3/5*1/27=1/x = 1/ 45= x= 45the 4th proportion = 45

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(1+sin x)/√(1-sin x) × √(1+sin x) /√(1+sin x)

= (1+ sin x )/√(1-sin²x) = (1+ sin x )/cos x

= ( 1/cos x +sin x/cos x ) = ( sec x + tan x ).

nth term of series = 3^nFirst term a1 = 3^1 = 3Second term a2 = 3^2 = 9

Common difference d = 9 - 3 = 6

Sum of n terms of series= n/2[2a1 + (n-1)d]= n/2[2*3 + (n-1)6]= n/2(6 + 6n - 6)= 3n^2

If 15ⁿ end with digit zero,then the number should be divisible by 2 and 5.

As 2×5=10

→This means the prime factorization of 15ⁿ should contain prime factors 2 and 5.

→15ⁿ = (3×5)ⁿ

It does not have the prime factor 2 but have 3 and 5,

Since 2 is not present in the prime factorization, there is no natural number nor which 15ⁿ ends with digit zero.

So,15ⁿ can not end with digit zero.

Let sum of n terms = 42We know sum of n terms of an APSn = n/2(2a1 + (n-1)d)

Given a1=2, d = 4-2 =2, Sn = 42

Then, 42 = n/2(2*2 + (n-1)2)42 = n (2 + n - 1)42 = n(n +1)6*7 = n(n+1)

Therefore n = 6Hence sum of 6 terms = 42

there would be one empty set ( 8)

8 mean thitha .thank you

b is the answer2025+1936-1369= 2592

the correct answer is finite

the correct answer is finite

it is finite set as we can count the total no. of states of tamil nadu

(a1/a2)=(b1/b2)=1/2

Hence given set of equations are inconsistent and uniques solution doesn't exist.

SetA=(1,2,3,4 ,5)=1024 =(1,2,4)

-2{-30}× -7-(-2)(-5)60× -7-10-420-10-430

0+1*6=6

6+3*6=24

24+6*6=60

60+10*6=120

120+15*6=210

210+21*6=336

sum of angles around a point is 360°then x+y+z+w=360°also x+y=w+zso x+y+x+y=3602(x+y)=360x+y=180since sum of angles on line AOB is 180° so it is a st. line

15SP=16CPso SP=16/15CPthat means it is gainso profit=(16/15CP-CP)/CP*100=100/15=6.67%

tan (A+B-C)= tan π/4sec(B+C-A)= π/3A+B-C= π/4B+C-A= π/3adding both2B= 105B= 52.5°

This is a Geometric Progression with common ratio (-1/2). So, a*r^n = 1/127=> r^n = 1/508.=> (-1/2)^n = 508n does not have any integer value for such an equation. So, this term does not exist for the progression.

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we have to find:-the sum of4-1/n, 4-2/n, 4-3/n = ?

solution:-we know that :1 + 2 +3 + 4 +5 +6 + ...........+ n = n(n +1) / 2and1 + 1+ 1 + 1 + 1 + .............+ n = n

Here,

sum of 4-1/n, 4-2/n, 4-3/n up to the nth term

= (4 + 4 + 4 + 4 + 4 + ......... upto n terms) + (-1/n - 2/n - 3/n - ..........upto n terms)

= 4 ( 1+1+1+1.......... upto n terms) - 1/n (1 + 2 + 3 +4 .........upto n terms)

= 4 n - 1/n× n(n +1)/2

= 4n - (n+1)/2

= [ 8n - (n+1) ] / 2 .......taking L.C.M

=( 7n - 1) / 2 Answer

possible subsets are{1},{2}, {1,2},{phi}

In this example, there are55elements, so there are2 power 5 = 32different subsets.

Here they are:

{}

{1}

{2}

{1,2}

{3}

{1,3}

{2,3}

{1,2,3}

{4}

{1,4}

{2,4}

{1,2,4}

{3,4}

{1,3,4}

{2,3,4}

{1,2,3,4}

{5}

{1,5}

{2,5}

{1,2,5}

{3,5}

{1,3,5}

{2,3,5}

{1,2,3,5}

{4,5}

{1,4,5}

{2,4,5}

{1,2,4,5}

{3,4,5}

{1,3,4,5}

{2,3,4,5}

{1,2,3,4,5}

Subsets of A ={1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}

Power set of A = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}}

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