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lawyer Dorey, Patricia K. Reviews

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I hope this is helpful. You have a resume with a "gap", and those are tricky. However, you have taken up education again, so that shows you are back on track - professionally speaking.

Health Care Admin University in Texas?

I have a project that's due in 2 weeks, and I'm freaking out!. I need to do a software about a loyalty program that has some of these functional requirements:. - Add points to customers based on their amount of purchase. (based on a particular business equation). - Deduct points when exchanged with a product. . Can I do such things using Oracle form Builder? Or should I use another programming tool.

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First, standard form allows us to write the equations for vertical lines, which is not possible in slope-intercept form. Remember that vertical lines have an undefined slope (which is why we can not write them in slope-intercept form). However, the vertical line through the point (4,7) has the standard form equation. 1 x + 0 y = 4. which we could write in the even more simple form. x = 4. [Note that the horizontal line through the point (4,7) has the slope-intercept form y=0x+7, and the standard form 0x+1y=7. This example demonstrates why we ask for the leading coefficient of x to be "non-negative" instead of asking for it to be "positive". For horizontal lines, that coefficient of x must be zero.]. A second reason for putting equations into standard form is that it allows us to employ a technique for solving systems of linear equations. This topic will not be covered until later in the course so we do not need standard form at this point. However it will become quite useful later.. . A third reason to use standard form is that it simplifies finding parallel and perpendicular lines. Let us look at the typical parallel line problem. Find the equation of the line that is parallel to the line 3x+4y=17 and that contains the point (2,8). The usual approach to this problem is to find the slope of the given line and then to use that slope along with the given point in the point-slope form for a linear equation. However, if we look at the standard form of a linear equation,. . Ax + By = C. and we move the Ax term to the other side. By = ? A x + C. and we divide both sides by B, assuming B is not zero, we get. y = (? A/B) x + C/B. which is the slope-intercept form. From that form we see that the slope is ? A/B. Any line parallel to the given line must have that same slope. Of course, the only values affecting the slope are A and B from the original standard form. Therefore, as long as A and B do not change, any line that has a standard form of. Ax + By = H. will be parallel to the line. Ax + By = C. If we return to the original problem, "Find the equation of the line that is parallel to the line 3x+4y=17 and that contains the point (2,8)" we can see that the answer must look like. 3x + 4y = H. and we just need to find the value of H. Of course, we also know that the point (2,8) must make the equation true, so. 3(2) + 48 = H. must be true. But this means that we have. 6 + 32 = H. or. 38 = H. Because we know the value of H, we have the complete answer. 3x + 4y = 38

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