Let's say 2a, 2a + 2, 2a + 4 are three consecutive even numbers 1. Show that the sum of the squares of these numbers can be described by phenomenon 4 (3a2 + 6a + 5);
2. Determine what these three numbers are if the sum of their squares is known to be 200 and the sum of these numbers is 24.

9514 1404 393

Step-by-step explanation:

1. The sum of squares is ...

  (2a)^2 +(2a +2)^2 + (2a +4)^2 = 4a^2 +(4a^2 +8a +4) +(4a^2 +16a +16)

  = 12a^2 +24a +20 . . . . . collect terms

  = 4(3a^2 +6a +5) . . . . . . factor out 4

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2. If the sum of the numbers is 24, we have ...

  2a +(2a +2) +(2a +4) = 24

  6a +6 = 24 . . . . . collect terms

  a +1 = 4 . . . . . . . .divide by 6

  a = 3 . . . . . . . subtract 1

  2a = 6

  2a +2 = 8

  2a +4 = 10

The numbers are 6, 8, 10.

Check

  The sum of squares is ...

  4((a +2)(3a) +5) = 4(5(3·3) +5) = 4(50) = 200 . . . . as required

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,