what work? If the smallest is x, then the next two are x+2 and x+4. So,
x^2 + (x+2)^2 = 54+20(x+4)
Now just crank it out.
Three consecutive odd intergers are such that the sum of the squares of the first two integers is 54 more than 20 times the third integer. Determine the three integers.
Please help and show all work thank you
SHOW WORK SO I UNDERSTAND HOW TO DO IT THANK YOU
1 answer
1 answer