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 Post subject: Reminder of how to solve these problem
PostPosted: Thu, 14 Oct 2010 15:25:06 UTC 
Hello Everyone , i would really appreciate any of you to remind me of how to solve some problems , i really need to understand how it goes on as ive already have started universty .

First one is : Find the equation of the circle with centre (-2,3) that passes through (1,-1).

Second is : Find which values of x satisfy the following , 1/2-x<2

3. Express the following as either open or closed intervals
{x: |x-4|≤ 6}

4. Express the following intervals in the form { x: | ax + b | < c } or { x: | ax + b | ≤ c }

(-4,-2)


Thank you


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 Post subject: Re: Reminder of how to solve these problem
PostPosted: Thu, 14 Oct 2010 21:21:54 UTC 
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Joined: Fri, 10 Oct 2008 15:35:23 UTC
Posts: 179
Location: Clarksville, ARkansas
Alghassab wrote:
Hello Everyone , i would really appreciate any of you to remind me of how to solve some problems , i really need to understand how it goes on as ive already have started university .

First one is : Find the equation of the circle with centre (-2,3) that passes through (1,-1).

Second is : Find which values of x satisfy the following , 1/2-x<2 .

3. Express the following as either open or closed intervals
{x: |x-4|≤ 6}

4. Express the following intervals in the form { x: | ax + b | < c } or { x: | ax + b | ≤ c }

(-4,-2)






First one: Find the equation of the circle with centre (-2,3) that passes through (1,-1).
The distance from the center, (-2,3), of the circle to the point (1,-1), which is on the circle, is equal to the radius of the circle. This gives:
r=\sqrt{\left(1-(-2)\right)^2+\left(-1-3\right)^2}.
Similarly, the distance from the center to any point, (x,\ y), on the circle is equal to the radius. Thus,
r=\sqrt{\left(x-1\right)^2+\left(y-(-1)\right)^2}. Square both sides to get the usual form for the equation of a circle.

r^2={\left(x-1\right)^2+\left(y-(-1)\right)^2

Of course, you will have a numerical value for r^2.


Second one: Find which values of x satisfy the following, \frac{1}{2}-x<2 .

Method 1: Subtracting 1/2 from both sides gives: \displaystyle{ -x<\frac{3}{2} }. Think about this: If the opposite (i.e. negative) of x is less than 3/2, then x itself must be greater than the opposite of 3/2.

Method 2: Adding x to both sides and subtracting 2 from both sides gives: \displaystyle{-\frac{3}{2}<x}. You can leave it like this or reverse the whole statement, including the sense (direction) of the inequality.

3. Express the following as either open or closed intervals. \{x: |x-4| \le 6\}
The quantity, |x-4| gives the distance that an arbitrary point x is from 4 on the number line. Your solution should be all the points which are a distance of 6 or less from 4. That should give you the answer.

To do this symbolically, think about what you would look for if you substitute various values in for x to get an answer by trial & error. You would need the absolute value of x-4 to be less than or equal to 6.
|x-4|\le6 That means the absolute value of x-4 needs to be between -6 & 6 or equal to -6 or 6. Writing this symbolically, we have:
-6 \le x-4 \le 6.
Add 4 to the left right AND middle of this inequality to get your result. Now convert your answer to interval notation.

4. Express the following interval in the form { x: | ax + b | < c } or { x: | ax + b | ≤ c } : (-4,-2)

Note: The parentheses, (), mean that x \ne -4 and x \ne -2.

The interval (-4,-2) is the set of points on the number line between -4 and -2 . This interval can be described as the set of points that are a distance of less than 1 from the point -3. Any point, x, on this interval must fulfill the condition:|x-(-3)|<1 .

Hopefully, you can take it from there.



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