Two consecutive positive numbers are such that the sum of their squares is 113. find the two numbers

7 and 8

Step-by-step explanation:

Let x represent the first number, then x + 1 is the other number.

(x)² + (x + 1)² = 113

x²  +  x² + 2x + 1 = 113    expanded (x + 1)²

2x² + 2x + 1 = 113          added like terms

2x² + 2x - 112 = 0        subtracted 113 from both sides

x² + x - 56 = 0           divided both sides by 2

(x + 8) (x - 7) = 0        factored polynomial

x + 8 = 0   x - 7 = 0    applied zero product property

x = -8     x = 7            solved for x

  ↓

not valid since the restriction is that x > 0 (a positive number)

So, x = 7     and     x + 1 = (7) + 1   = 8

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72