Three consecutive natural numbers ae such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60

The three consecutive natural numbers are 9,10,11

Step-by-step explanation:

Step 1 : -

Let x be the number

Given three consecutive natural numbers are x, x+1,x+2

Given data are three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60

that is [tex](x+1)^2 = (x+2)^2 - x^2 +60{\tex]

step 2:-

on simplification on both sides are , we get

by using [tex](a+b)^2 = a^2+2 a b+b^2[\tex]

[tex]x^2+2 x+1 = x^2+4 x+4-x^2+60[\tex]

cancelling x^2 terms and simplify

[tex] x^2 -2 x-63=0[\tex]

now finding factors of [tex] 63 = 9 X 7[\tex]

[tex] x^2 - 9 x+7 x -63=0 [\tex]

[tex] x(x-9)+7(x-90 =0 [\tex]

[tex]( x+7)(x-9) =0 [\tex]

here x= -7 is not an natural number

so we have to take x=9

therefore the three consecutive natural numbers are

x,x+1,x+2

The three consecutive natural numbers are  9 , 10, 11

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.

∡ eia and ∡ kij are congruent

step-by-step explanation: