Find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18?

Let, numbers = x,(x+2),(x+4),(x+6)

A.T.Q,

x+x+2+x+4 = x+6+18

3x+6 = x+24

3x-x= 24-6

2x = 18

x = 9

Numbers would be 9, 11, 13, 15

9,11,13,15

Step-by-step explanation:

Let our 4 numbers be

-n

-n+2

-n+4

-n+6

first, we have to write an expression and we gradually simplify.

n+n+n+2+4=n+6+18

combine like terms:

3n+6==n+24

subtract from both parts of the equation.

3n-n=24-6

now are finally finishing our problem.

2n=18

n=9

Now that we know our first integer, we need to consecute.

9,11,13,15

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

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step-by-step explanation:

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