Find three consecutive odd intergers whose sum is 219

1st odd = 71

2nd odd  = 73

3rd odd  = 75

Step-by-step explanation:

As numbers are consecutive odd so let:

1st odd = x

2nd odd = x + 2

3rd odd = x + 4

According to given condition:

x + x +2 + x + 4 = 219

Adding like terms

3x + 6 = 219

Subtracting 6 from both sides

3x + 6 -6 = 219-6

3x = 213

Dividing both sides by 3

x = 213/3

x = 71

So the numbers will be:

1st odd = x = 71

2nd odd = x + 2 = 71 + 2 = 73

3rd odd = x + 4 = 71 + 4 = 75

Proof:

1st odd + 2nd odd + 3rd odd = 219

71 + 73 + 75 = 219

219 = 219

Hence proved!

I hope it will help you!

rounding to the nearest thousandth would be 452,000

m a t h w a y. c o m

step-by-step explanation: