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Given:- the cubical box has a volume of 64 cubic cm.

Let the cubical box has an edge of side 'a'.=>a³=64=>a=4 cm

Option A is correct.

lcm of 2,3,4,5=60

Now, using 60 as first term and 960 (above 960 and below 1000 there is no other number which is divisible by 60)as nth term and difference (d)=60 ,

We can calculate ‘n' value

tn=a+(n-1)d

960=60+(n-1)60

16=1+(n-1)

n=16.

We can say that there are 16 numbers which are divisible by 2,3,4 and 5.

Volume of cube=64 cm³(edge)³=64 cm³edge= 4cm

boys are 28 and girls are 20 so total are 48

how

Let number of boys = 7xLet number of girls = 5x

As per given condition7x = 5x + 82x = 8x = 8/2 = 4

Total class strength= 12x = 12*4= 48

V180=V90/2=V45/4=V9x5/4=3V5/4

The decimal form is 9.025

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a pahala wala questions

Below are some of the Standard Algebraic Identities:

Identity I:(a + b)2= a2+ 2ab + b2

Identity II:(a – b)2= a2– 2ab + b2

Identity III:a2– b2= (a + b)(a – b)

Identity IV:(x + a)(x + b) = x2+ (a + b) x + ab

Identity V:(a + b + c)2= a2+ b2+ c2+ 2ab + 2bc + 2ca

Identity VI:(a + b)3= a3+ b3+ 3ab (a + b)

Identity VII:(a – b)3= a3– b3– 3ab (a – b)

Identity VIII:a3+ b3+ c3– 3abc= (a + b + c)(a2+ b2+ c2– ab – bc – ca

Answer:d) 1/2Explanation:No of outcome= 6even numbers=3(2,4,6)now,P(even numbers)=No of possible outcome /total outcome=3/6=1/2

total events=6 and no of an even no. =3 possibility of getting an even number=no. of outcome÷ no. of total events=3÷6=1÷2

Probability = possible outcomes /total outcomes

When dice is rolled total outcomes = 6

(1) Possible outcomes of evenNumber = 3Probability of even number= 3/6 = 1/2

(2)Possible outcomes of oddNumber = 3Probability of odd number= 3/6 = 1/2

Putting n=1, 5n²-3n=5-3=2

Putting n=2, 5n²-3n=14

Therefore 2nd term is 14-2=12

Therefore AP is 2, 12, 22 ... and so on.

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Given:Sn = 2n²+3nan = S(n) - S(n-1)=2n²+3n -[ 2(n-1)²+3(n-1)]= 2n²+3n -[2(n²+1-2n)+3n -3]= 2n²+3n -[2n²+2-4n + 3n -3]= 2n²+3n -2n²+4n-3n -2+3= 2n²+2n²+3n -3n +4n -2+3= 4n +1an = 4n+1a16= 4×16 +1=64+1= 65Hence the 16th term is 65

nth term=5n^2-3nlet n=1.,=5-3=2n=2, 5(2)^2-3*2=20-6=14n=3,. 5(3)^2-3*3=45-9=36n=10. ,5(10)^2-3*10=500+30=570

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First Term, a=7-3*(1)=4Second term=7-3*(2)=1d=1-4=-3

Sum of first 25 terms=( 25/2)*((2*4)+(25-1)*(-3))=(25/2)*(8-72)=-25*32= -800

1/cos²theta+1/sin²thetasin²theta+ cos² theta/cos²theta.sin²thetaas we know sin²theta+ cos²theta= 1so 1/sin²theta.cos²theta = sec² theta. cosec²theta

oh that's too easy I have done many simplification for this question but it's simple

thanks bro

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sum of p terms=p (p+1)(p+2)S1 be sum of 1 termS1=a1=1 (1+1)(1+2)=6S2 be sum of 2 termsS2=a1+a2=2 (2+1)(2+2)=24

a2=24-a1=24-6=18

d=a2-a1=18-6=12a10=a+9d=6+9×12=6+108=114a10=114

thanks

draw a line parallel to P , through O ,

so angle made by the line with the lower O + 120° = 180°

=> angle O = 180-120 = 60°

now angle 60+30 + x = 180° .....( corresponding angle)

=> x = 180-60-30 = 90°

Thanks a lot for helping me

you are welcome, anytime

The pair of linear equations represented by these lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

If a1/a2 =/= b1/b2 then the pair of linear equations has exactly one solution.

Therefore,For equations kx - y = 2 and 6x - 2y = 3 to have unique solution

k/6 =/= - 1/-2k =/= 3

A2/3 is right answer

5xyz^2-3zy=yz(5xz-3)

5xyz*z - 3yz = yz (5xz - 3)

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terms are 3a , -7b and 6z factors are 3 and a (-1),7 and b 6 and z

a) 1960 of this question

5 सन्ख्या का औसत 15 योग = 75 4 संख्या 18 9 21 16 पांचवी संख्या = 75-18-9-21-16 = 11