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Slope of A and B is

0 - (-2)---------- = 2/7x - (-3) 2---------- = 2/7 x + 3

7 × 2 = 2 ( x + 3)

7 = x + 3

x = 7 - 3

x = 4

(a) option is correct

Part 1

thanks

x²+7x+12=x²+3x+4x+12=x(x+3)+4(x+3)=(x+4)(x+3)

Thank u

x^2*d/dx(cosx)+cosx*d/dx(x^2)=-x^2*sinx+2xcosx

does the denominator contains only cosx ?? or is it like this... cosx/(cosx+4sinx) .?

yes find it's integral

cosx divide by cosx +4sinx

If this point lies on the line, it's coordinates should satisfy the equation of the line. 3(p)+2 = 11=> 3p = 9=> p = 3.

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Here a= 2 and b= 3√5 and c= 5so D= b^2-4ac= 45-(40)= 5now we know that roots = -b(+-)√D/(2a)so roots will be -3√5(+-)√5/4

2x² + 3√5x + 5

=> 2x² + 2√5 x+ √5x+5 {< by using splinting method )

=> 2x( x +√5 ) + √5 (x +,√5)

=> (2x +√5)(x+√5) =0

=> x =-√5/2 and x =- √5 Answers ..

Thanks

4+3=5x=> 5x=7=> x=7/5

2x² -5x + 2 =02x² - 4x -x + 2 = 02x(x - 2) - (x - 2) = 0(2x - 1) ( x-2)=0x = 2, x = 1/2

Thanks

LHS = ( cosA-sinA+1)/(cosA+sinA-1)

divide numerator and denominator with

sinA , we get

= ( cotA - 1 + cosecA)/( cotA+1 - cosecA )

= [cotA+cosecA- 1 ]/[ 1 - cosecA + cotA ]

=[(cotA+cosecA)-(cosec²A-cot²A)]/(1-cosecA+cotA)

=[(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]/(1-cosecA+cotA)

= [(cosecA+cotA)(1-cosecA+cotA)]/(1-cosecA+cotA)

= cosecA + cotA

= RHS

(-8) × ? = 504 ? = 504/(-8) = -63

The prime factors are: 3 x 11 x 13

or also written as { 3, 11, 13 }

Written in exponential form: 3^1x 11^1x 13^1

i) 3x² = 2x+8 => 3x²-2x-8 = 0 => 3x²-6x+4x-8 = 0 => 3x(x-2)+4(x-2) = 0 => (3x+4)(x-2) = 0 , the value of x = x = -4/3 , 2

ii) 2x²+5x+2 = 0 => 2x²+4x+x+2 =0 => 2x(x+2) +1(x+2) = 0 => (2x+1)(x+2) =0 so the value of x = -1/2 , -2

iii) x²-5x-36 = 0 => x²-9x+4x-36 =0 => x(x-9)+4(x-9) =. => (x+4)(x-9) = 0 so x = 9,-4.

iv) (2x+3)² = 16 => 2x+3 = √16 = +4,-4 => 2x = +4-3 , -4-3 => x = 1/2 , -7/2.

d/dx(sin2x)

= d/dx(2sinxcosx)

Use product rule of

= 2[sinx d/dx(cosx) + cosx d/dx(sinx)]

= 2[sinx(−sinx) + cosx(cosx)]

= 2(cos^2x−sin^2x)

Trigonometric identity

cos^2x−sin^2x = cos2x

= 2(cos2x)

4x^2+12x+5when we add 4 in the equation it will become4x^2+12x+5+4=(2x)^2+2(2x)(3)+(3)^2=(2x+3)^2

I didn't uderstood

solve it by shreedharacharya formula

solve completing square method

If 12ⁿ ends with 0 then it must have 5 as a factor.But, 12ⁿ=(2×2×3)ⁿ which shows that only 2 and 3 are the prime factors of 12ⁿ.Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.So, 5 is not a factor of 12ⁿ.Hence, 12ⁿ can never end with the digit 0.

let no.s be x and yx+y=25/2------(1)x-y=25/12-----(2)2x=25/2 + 25/12=175/12x=175/24put in eq 2175/24-y=25/12y=175/24-25/12y=75/24y=25/8

(x/3)+3/(6-x)=2(6-x)/15; (6-x)x+9/3(6-x)=12-2x/15; 6x-x^2+9/18-3x=12-2x/15; 15(6x-x^2+9)=12-2x(18-3x); 90x-15x^2+135=216-36x-+36x+6x^2;;216-135-9x^2+90x=81-9x^2+90x=9(9-x^2+10x)

x(6-x)+9=12+2x/15; 6x-x^2+9=12+2x/15; 15(6x-x^2+9)=12+2x ; 90x-15x^2+135=12+2x; 88x-15x^2+123

x=2 is the right answer

It means that he had bought 10 ice creams for ₹75now 1 ice cream for ₹75/10So he purchased total ice cream =100÷7.5=1000÷75=13. approximately

1 is 100% of 13 3so 2 is twice the 1 3 3and 1 is 100% 3so 2 is 200% 3the ans is d 200%

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