Two numbers are such that one number is 42 more that the second number and their average is equal to 40. what is the larger of the two numbers?

The larger number be 61.

Step-by-step explanation:

Let the second number be 'x'.

Let the first number be 42+x

Average of two numbers = 40

According to question,

So, the larger number of the two numbers i.e the first number is

Hence, the larger number be 61.

largest of the two number is 61

Step-by-step explanation:

given,

one number is 42 more that the second number

and average is equal to 40.

let as assume first number be x

so the second no becomes = x + 42

average of the two numbers = 40

so the first number is 19

second number will be equal to 19 + 42 = 61

hence the largest of the two number is 61

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

(x+3)(x+7)

step-by-step explanation: