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1. (5i)(-3i/5) = -i²3 = 3

(3x +1) - 7 = 123x - 6 = 123x = 12 + 63x = 18x = 6 ans

3x+2=113x=11-23x=9x=9÷3x=3

3x+2=113x=11-23x=9x=9÷3x=3

3x +2 =113x=11_23×=9×=9÷3×=3

3x+2=113x=11-23x=9x=9/3=3

2 tan ^-1( cosx)= tan^-1( 2cosecx); solving 2 tan^-1( cosx); we know that 2 tan^-1x= tan^-1 2x/1-x^2; replacing x with cosx 2 tan^-1( cosx)=tan^-1 2cosx/ 1- cosx^2= = tan^-1(2cosx/ sinx); 2 tan^-1( cosx)= = tan^-1(2cos x/ sinx^2) =2 cosx/ sinx^2=2 cosecx cosx/ sinx^2= cosecx cosx/ sinx^2=1/ sinx cosx= sinx^2/sinx = cosx= sinx sinx/ cosx=1

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3x - 5 = 5x + 113x - 5x = 11 + 5-2x = 16x = 16/(-2) x = -8

The shaded region contains all the points lying in the equation 3x+2y > 6

It has equal roots when D=0 (b) square-4ac= (2√5)sq-4(p)(15) =020-60p=0 p=1/3.

what's your what's app no.

28x^4/56x= 1/2 x^4-1= 1/2 x^3

suppose x per cent of 2/7 is 1/35.so (x*2/7)/100=1/35so x=(1/35)*(7/2)*(100)=10so 10 per cent of 2/7 is 1/35

Let the percentage be xx=(1/35)×100/(2/7)=100×7 /35×2 =10%

if a is a root, then f(a) = 0

by inspection of the coefficients (they add to 0), then 1 is a root and x - 1 is a factor(note that f(1) = 2 + 1 - 5 + 2 = 0

divide either synthetically or by long division to get:

1 | ... 2 .... 1 ..... -5 ..... 2...... ......... 2..... 3 .....-2...---------------- ---------------- (add)....... 2...... 3 ......-2 .....0 <== remainder is 0, confirming that x - 1 is a factor

quotient is 2x^2 + 3x - 2, which factors as (2x - 1)(x + 2)

thus, fully factored this is f(x) = 2x^3 + x^2 - 5x + 2 = (x - 1)(2x - 1)(x + 2)

................x^2 ..+ 2x .+ 1/2.........------------------------------...2x - 3 | ... 2x^3 + x^2 - 5x + 2...............2x^3 - 3x^2............. ------------------ (subtract)........................ 4x^2 - 5x....................... 4x^2 - 6x........................ ------------- (subtract)...................................x + 2........................... .......x - 1.5.................................. ---------............................ ........... 7/2 <== remainder

(2x^3 + x^2 - 5x + 2) / (2x - 3) = x^2 + 2x + 1/2 + (7/2) / (2x - 3)

or, 2x^3 + x^2 - 5x + 2 = (x^2 + 2x + 1/2)(2x - 3) + 7/2

just looking at the constants: (1/2)(-3) + 7/2 = -3/2 + 7/2 = 2 as needed

remainder = 7/2

The nth term is an=3+2n so by using this we can find the value of terms like a1=5,a2=7,a3=9and a24=51now the first term is 5 and the last term is 51 and the difference is 2by using the formula for sum of n terms i. esn =n/2(2a+(n-1)d)so s24=24/2(10+23*2) i. e 672 ans.

x=1 is your correct answer

14x+14x=2828x=28x=28/28=1

Normal hematopoiesis is a well-regulated process in which the generation of mature blood elements occurs from a primitive pluripotent stem cell in an ordered sequence of maturation and proliferation. Regulation occurs at the level of the structured microenvironment (stroma), via cell-cell interactions and by way of the generation of specific hormones and cytokines: erythropoietin, interleukin 3, granulocyte-monocyte colony-stimulating factor (GM-CSF), monocyte-macrophage colony-stimulating factor (M-CSF), granulocyte colony-stimulating factor (G-CSF), interleukin 5, interleukin 4, and other less well-defined factors, including the megakaryocyte growth factors. Understanding of this complex process has revealed insights into the pathophysiology of human disease and provided a theoretical framework for the therapeutic use of bone marrow transplantation and potential gene transfer therapy. Furthermore, ongoing clinical trials suggest that the hematopoietic growth factors may represent a significant new group of therapeutic reagents for patients with hematological and oncologic disease

2cos^2x+3sinx=02(1-sin^2x)+3sinx=02-2sin^2x +3sinx=02sin^2x -3sinx -2=02sin^2x -4sinx+sinx-2=02sin x(sin x-2)+1(sinx-2)=0(sin x -2)(2sin x+1)=0sin x - 2=0 or 2 sin x+1=0sin x =2 or sin x = -1/2

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consider the value of gx is equal to zero then what ever the value comes after equating gx with zero put that value in place of x in p of x

Given that, mth term=1/n and nth term=1/m.then, let a and d be the first term and the common difference of the A.P.so, a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2)subtracting equation (1) by (2) we get,md-d-nd+d=1/n-1/m=>d(m-n)=m-n/mn=>d=1/mn. again if we put this value in equation (1) or (2) we get, a=1/mn.then, let A be the mnth term of the APa+(mn-1)d=1/mn+1+(-1/mn)=1 hence proved.

put x = (-1))remainder will be

thanks

(x - 1) is a factor of P(x) = x⁴ - 2x³ + 3x - 2

means, x = 1 is a root of P(x) = x⁴ - 2x³ + 3x - 2.

P(1) = 1⁴ - 2(1)³ + 3(1) - 2

P(1) = 1 - 2 + 3 - 2

P(1) = 4 - 4 = 0

we get, P(1) = 0 hence, the given statement is correct that (x - 1) is a factor of p(x)=x⁴–2x³+3x–2.

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nth term in AP=a+(n-1)d where a is first term and d is common difference

The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on.

Thus nth term of an APseriesis Tn= a + (n - 1) d, where Tn= nthterm and a = first term.

Here d = common difference = Tn- Tn-1.

The sum of n terms is also equal to the formula where l is the last term.

2cos^2x+3sinx=02(1-sin^2x)+3sinx=02-2sin^2x +3sinx=02sin^2x -3sinx -2=02sin^2x -4sinx+sinx-2=02sin x(sin x-2)+1(sinx-2)=0(sin x -2)(2sin x+1)=0sin x - 2=0 or 2 sin x+1=0sin x =2 or sin x = -1/2

100 paise=1 rupeeshence 104 rupees =100*104=10400paisehence total paiseis 10400+25=10425paise

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Let a, b and c are zeroes of given polynomial 2x^3 + 5x^2 - 4x + 3 = 0.

Then multiplication of all zeroes (a*b*c) = - constant term/ coefficient of x^3= - 3/2

Area=a^2d=√2 ad=√(2A)Area=676cm^2a^2=676cm^2a=√676a=26d=√2×26=1.414×26=36.764cm

let the C matrix be [ x,y,z ] [u,v,w]

For A+B+C = 0. all the sum of elements should be zero. 1+2+x =0 => x = -3. similarly y = 4 , z = -1 u = -3 , v = 0 , and w = -1

so the matrix C is first row [-3,4,-1] 2nd row [-3,0,-1]

i) 4, 16, 64

pattern is 4ⁿSo, First term = 4Second term = 4² = 16n th term = 4ⁿ

ii) 10, 20, 30, 40.....

Pattern is 10, 10 + 10, 10 + 10 + 10....

So, n th term is 10n

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Cube shaped tin length = 30cmVolume of cube = 6*length*length = 6*30*30 = 5400 cm^3

Oil can be filled in tin = 5400*.001 = 5.4 litres

Rs. 5 = 10* 50 paise

7m-(-5)(2)=317m+10=317m=31-107m=21m=21/7=3Therefore m=3

The number will be 20-e