Question

pls tell the answer

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Answer 1

Step-by-step explanation:

Solution:

Let the two consecutive Numbers be x and x+1.

Therefore ,

x² + (x+1)² = 365

x² + x² +1² + 2*x*1 = 365 (because (A+B)² = A² + B²+ 2AB)

or 2x² + 1 + 2x = 365

2x² + 2x = 365 - 1

2x² + 2x = 364

2(x² + x) = 364

or x² + x = 364/2

or x² + x = 182

or x² + x - 182 =0

Now Solve the Quadratic Equation ,

x² + 14x - 13x - 182 = 0

Note : - 13 *14 = 182 , this is because I write 14x - 13x instead of x , so as to solve the quadratic equation .

x (x+14) - 13 (x +14 ) = 0

( x- 13 )(x+14)=0

Therefore , Either x - 13 = 0 or x+14 =0

Since the Consecutive Integers are positive ,

therefore , x-13 = 0

⇒ x =13

hence One of the Positive Integers = 13 ,

therefore other positive integer = x+1 = 13+1 = 14

So the two consecutive positive Integers are 13 and 14 .

Hope this helps You !!

Thanks Cheers !!

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

$\implies \: 12 : 3 : :x : 1 \\ \implies \: \frac{12}{3} = \frac{x}{1} \\ \implies3 \times x = 12 \times 1 \\ \implies \: x = \frac{ \cancel12}{ \cancel3} \\ \implies \boxed{x = 4}$

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \#\mathcal{answer with quality }\: \: \& \: \: \#BAL }$