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whats your question ?
think you can help?
Since you want to consider 2 consecutive EVEN integers, you conceive of the first one as: 2x and the next one as 2x + 2 The multiplier of 2 guarantees that it will be even: 1/(2x) + 1/(2x + 2) = 9/40
1/x + 1/(x + 1) = 9/20 [(x + 1) + x] / [x(x + 1)] = 9/20 40x + 20 = 9x^2 + 9x 9x^2 - 31x - 20 = 0 x = 4 -> so 2x = 8 and the next # is 10 1/8 + 1/10 = 9/40
im kinda confused as to what ll of this means
You're looking for 2 consecutive even numbers such that 1 over the first number plus 1 over the second number = 9/40 Those 2 numbers are 8 and 10 because: 1/8 + 1/10 = 9/40 And the equation is set up with the first number being represented as: 2x and the second # is then 2x + 2 The "2" will guarantee that your eventual numbers are even. So you start with that, go to the equation I set up, and solve for "x". Then you take 2x for the first # and 2x + 2 for the second #.
You don't have to find anything. The problem is started and finished. The first number is 8 and the second number is 10.
The reciprocal of 8 is 1/8 The reciprocal of 10 is 1/10 1/8 + 1/10 = 9/40 The 2 numbers that answer the question are 8 and 10.
Well, the answer is in 2 parts. You need the initial equation and then you need: 8 and 10.
Actually, the equation is: 1/(2x) + 1/(2x + 2) = 9/40
The problem is asking for an equation than can be used to find the 2 numbers. So, the equation has to have a variable in it.
I suggest you go over all the logic here. It might help.
yeah that's what im doing!
ok, well thx for the recognition and have a great day.