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let alpha= p bita = qx^2+6x+9p+q= -6/1= -6pq= 9/1=9sum of given zeros+p+(-q)= -(p+q)= -(-6)=6product of given zeros (-p)(-q)= pq= 9so polynomial = x^2-x(sum of given zeros)+(product of given zerosx^2-x(6)+9x^2-6x+9

X²-6x+9 is the answer

taking tan^-1 to both the sides

if x = u+v/(1-uv)

=> tan^-1(x) = tan^(u+v)/(1-uv) = tan^-1(u) + tan^-1(v) similarly tan^-1(y) = tan^-1(u) - tan^-1(v)

now adding both tan^-1(x) + tan^-1(y) = tan^-1(x+y)/(1-xy) => tan^-1(u) +tan^-1(v) +tan^-1(u) -tan^-1(v) = tan^-1(x+y)/(1-xy)

=> 2tan^-1(u) = tan^-1(x+y)/(1-xy)=> tan^-1(2u/1-u²) = tan^-1(x+y)/(1-xy)

so, (x+y)/(1-xy) = 2u/(1-u²)

thanks bro for helping me

It is proper fraction

both are proper fraction

proper fraction .

your answer is coorect

Thank you!

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Euler Formula : V− E + F = 2

Givenvertices = 6 edges = 126 - 12 + F = 2F = 2 + 6 = 8there are 8 faces

and 1/2cos-1(3/5) = 1/2tan-1(4/3)

11th term=a+10d=3816th term=a+15d=73so 31st tem=a+30d=?a+15d-a-10d=73-385d=35so d=7so a+70=38so a=-32so a+30d=-32+30(7)=-32+210=178so 31st term is 178

11th=38; 16th=73; 11th= a + 10d =38, 16th=a + 15d=73/5d=35; 5d=35; d = 35/5=7; a=38-10d=38-10(7)=38-70=-32; a=32; n=31,;d=7

it is correct answer

a= 32 n= 31 d = 7 is the correct answer of the given

let 1st term = a , diff = d11 th term = 38a+10d= 38a+15d= 73by subtracting -5d= -35d= 35/5=7a+10d= 38a= 38-10×7 = 38-70= -32so 31 th term,= a+(n-1)d = -32+(31-1)×5 = -32+150 = 118

We know that,an=a+(n-1) dGiven :11 th term is 38a11=a+(11-1)d38=a+10d38-10d=aa=38-10d...(1)16 th term is 73a16=a+(16-1)d73=a+15d73-15d=aa=73-15d...(2)from (1) and (2)38-10d=73-15d38-73=-15d-10d-35= -5d-35÷-5=dd= 7Substitute value of d in equation (1)a=38-10(7)a=38-70a=-32we need to find 31st term,n=31,a=-32,d=7an=a+(n-1)da31=-32+(31-1)7 =-32+30×7 =-32+210 =178Therefore 31st term of AP is 178

So, from i) and ii) OP is the perpendicular bisector of AB

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please send the solution of full screen

3m² = 2m² - 93m² - 2m² + 9 = 0m² + 9 = 0Os herea = 1b = 0c = 9

ii) -10+3=-7; iii) -75+18=-57 iv)19-25=-6 (v)27-27=0 (vi)-20+0=-20, ( vii)-35-10=-45

Normally, year consists of 365 days .But leap year had 365+1 days .

366 days divided into weekly days has 365/7 i.e. 52 days + 2 extra days.

U have t find that two days are Sunday and Monday then we get the 53 Sundays or 53 Mondays.

It cleared that weekly days are cyclic.

Two days should be one of these sets >[(S,M),(M,T),(T,W),(W,TH),(TH,F),(F,Sa),(Sa,S)].

Therefore ,the probability that getting 53 Sundays and Mondays = the probability that 2 extra days should be sun or mon.

There are seven cases that the two days may be happen like that (already shown above).

In those, (S,M), (M,T) that has to be happen to get 53 sun or 53 mon days.

Therefore,Required probability = 2/7

(ii) is correct option

l only ask the sunday

57+58+x = 180now x = 65

123 + 122 + x = 180; =245 + x = 180; x = 180 - 245 =65

X = 65 degree this is the correct answer for this question Thank you

correct answer is 65

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65 Degree is correct answer

A) Sum of angles is 180 degree. Hence the pair of angles are supplementary.

B) Sum of angles is 90 degree. Hence the pair of angles are complementary.

C) Sum of angles is 90 degree. Hence the pair of angles are complementary.

D)Sum of angles is 180 degree. Hence the pair of angles are supplementary.

7 8 = 7×1 + 1 18 = 8×2 + 2 57 = 18×3 + 3 57×4 + 4 = 232 1165 = 232×5 + 5

232

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right

Given condition forms right triangle between two poleswith base b = 12 m, Height h = 11 - 6 = 5 m

Let distance between top of poles be x then,

Using pythagoras theoramx^2 = b^2 + h^2 x^2 = 12*12 + 5*5x^2 = 144 + 25x = 13

Therefore, distance between top of the poles = 13 m

y = x/2 3x + 4y = 20 3x + 4(x/2) = 20 3x + 2x = 20 5x = 20 x = 4 y = 4/2 = 2

A-wednesdayb-thursdayc-tuesday

write the procedure of calculation pliz

pliz do calculate

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Any odd positive integerncan be written in form of 4q+ 1 or 4q+ 3.Ifn= 4q+ 1, whenn2- 1 = (4q+ 1)2- 1 = 16q2+ 8q+ 1 - 1 = 8q(2q+ 1) which is divisible by 8.Ifn= 4q+ 3, whenn2- 1 = (4q+ 3)2- 1 = 16q2+ 24q+ 9 - 1 = 8(2q2+ 3q+ 1) which is divisible by 8.So, it is clear thatn2- 1 is divisible by 8, if n is an odd positive integer.