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waw gred very nice good your prssnn

put A = 7

3 7 + 2 5 -------------- 6 2

11x + 15y = - 23....(1)7x + 2y = 20..........(2)

Multiply eq(1) by 7 and eq(2) by 11

77x + 105y = - 161.....(3)77x + 22y = 220.........(4)

Subtract eq(4) from eq(3)83y = - 381y = - 381/83

Put value of y in eq(2)7x + 2*(-381)/83 = 207x = 20 + 762/837x = (1660 + 762)/837x = (2422)/83x = 346/83

In triangle ADC and ABC AB = DC AD = BC AC = AC by SSS rule ∆ADC is congurent to ∆ABC

m and n are zeroes of ax^2 - 5x + cGiven,m + n = 10 = mn

We know, Sum of zeroes = 5/a10 = 5/aa = 5/10 = 1/2

Product of zeroes = c/a10 = c/ac = 10*1/2c = 5

n= r+shere r is 8 and s is 2n=8+2=10

2y(-11x + 7) - (-5(xy - y)

= - 22xy + 14y + 5xy - 5y

= - 17xy + 9y

= 9y - 17xy

by basic multiplication(-6)^4=6^4=36^2=1296

2^6= 2*2*2*2*2*2= 64

(1)= 2x2x2x2x2x2=64

(2)=6^4hope this will help you like my answer and MARK IT as expert answer

2*2*2*2*2*2=642raised to power 6 means we will multiply 2 six times

2^6=2×2×2×2×2×2=64 correct answer.

2^6=2×2×2×2×2×2=64Answer

2×2×2×2×2×2=64 so 2 to the power of 6 is64

(2)^664 is correct answer

64 is answer okkkkkk

value of 2 will be 64

1.(i) 2*2*2*2*2*2 is my answer

=2*2*2*2*2*2*=64 ok ok ok

64 is the best answer

2*2*2*2*2*2* 64 ok answer

Let the numbers be 3x, 3x + 3 and 3x + 6.Then,3x + (3x + 3) + (3x + 6) = 729x = 63x = 7Largest number = 3x + 6 = 27.

=> Second largest number = 27 - 3 = 24

thanks

your answer is O=12A=8AB=3B=6total number is 30

your answer is 0=12A=8AB=3B=6total number is 30

Perimeter = 2( length+breadth)64/2= 17 + breadth32-17= breadth15 cm= breadthPlease like the solution 👍 ✔️👍

A= {1, 2, 3, 4, 5}

R = { (a,b) ; |a – b| is even}

It is clear that for any elementa∈A, we have |a -a| = 0(which is even).

∴R is reflexive.

Let (a,b) ∈ R.

=> |a –b| is even.

=> |- (a –b)| = |b - a| is also even.

=> (b, a) ∈ R is even.

A= {1, 2, 3, 4, 5}

R = { (a, b) : | a – b| is even}

It is clear that for any elementa∈A, we have |a - a | = 0(which is even).

∴R is reflexive.

Let (a,b) ∈ R.

⇒|a –b| is even.

⇒|- (a –b)| = |b - a| is also even.

⇒(b, a) ∈ R is even.

∴R is symmetric.

Now, let (a,b) ∈ R and (b,c) ∈ R.

⇒ |a –b| is even and |(b –c)| is even.

⇒ (a – b) is even and (b –c ) is even.

⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]

⇒ |a – c | is even.

⇒ (a,c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

GIVEN nc8= nc2 if ncr= ncpthen either r=p or r= n- p thus,nc8= nc2 8= n-2 8+2= n 10= ncalculating ,nc2 put n = 10 10 c2= 10/2(10-2) :::::::::::::::::::::::::::::::::::::→ncr = n / r( n- r)←:::::::::::::::::::::::::::::::::::::

GIVEN nc8= nc2 if NCR= ncPthen either r=p or r= n- p thus,nc8= nc2 8= n-2 8+2= n 10= ncalculating ,nc2 put n = 10 10 C2= 10/2(10-2) :::::::::::::::::::::::::::::::::::::→NCR = n / r( n- r)←:::::::::::::::::::::::::::::::::::::

The product of 11 and 6 is 66

answer already given below

15-2x+5y-11xy+2xy^2+8x^2y-(3x^2y-2xy+2xy^2+5x-7y-10)=15-2x+5y-11xy+2xy^2+8x^2y-3x^2y+2xy-2xy^2-5x+7y+10=25-7x+12y+5x^2y

6) suppose numbers are 11(x-1),11x,11(x+1)so 11x-11+11x+11x+11=363so 33x=363 so x=11so numbers are 0,121,1327) suppose numbers are 8(x-1),8x,8(x+1)so 8x-8+8x+8x+8=888so 24x=888so x=37so numbers are 288,296,304

x=1 is right answer.........

x=1

check:

6*1+4*1=10=6+4=10

tq

4x + 10 = 504x = 50 - 104x = 40x = 40/4x = 10