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This is my second post for what i got wrong in a test, and would like to be shown what i did wrong.

List the integers which satisfy both of the following inequalities

2x - 9 <0 and 8- x < 6 ( the < sign on this one is meant to be the less than or equal to however i cannot do it on the keyboard :) )


Also i have another one that confused me

The remainder when X(cubed) - 2X + 4 is devided by (x-2)is twice the remainder when X(squared)+X+K is devided by (x+1)

It wanted me to find the value of K, but whenever i tried working it out i kept getting the remainder as 0 , cant understand where i went wrong

2007-11-27 04:17:51 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

PROBLEM 1:

Solve each equation in terms of x:

First equation:
2x - 9 ≤ 0
2x ≤ 9
x ≤ 9/2

Second equation:
8 - x ≤ 6
8 - x - 6 ≤ 0
2 - x ≤ 0
2 ≤ x
x ≥ 2

Combining these:
2 ≤ x ≤ 4.5

So valid integer answers are {2, 3, 4}

PROBLEM 2:

Be sure to include a 0x² when you are dividing. Also be careful when you subtract negative terms. For example 4 - (-4) is 8, not 0.

..................... x² + 2x + 2
x - 2 ) x^3 + 0x² - 2x + 4
......... x^3 - 2x²
.................. 2x² - 2x
.................. 2x² - 4x
........................... 2x + 4
........................... 2x - 4
.................................. 8

So the remainder is 8 which is twice the remainder in the second equation. Let's divide the second equation:
.................. x + 0
x + 1 ) x² + x + K
.......... x² + x
................. 0 + K
..................0 + 0
........................ K

Since 8 is twice K...
2K = 8
K = ½(8)
K = 4

2007-11-27 04:33:28 · answer #1 · answered by Puzzling 7 · 0 0

2x - 9 < 0.
x - 4.5 < 0
If x is an integer, then it is any number between -∞ and less than or equal to 4.

8 - x < 6
Than -x < 6 - 8
-x < -2
x > 2
If x is an integer, then it is any number greater that 2 and less than ∞.

So the only integer that works for both inequalities are 3 and 4.

2007-11-27 12:26:03 · answer #2 · answered by Edgar Greenberg 5 · 0 0

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