Demonstrate the power of Microsoft Access queries by creating a query and modifying criteria. Add a field that contains the results of applying a formula to the records selected. http://sites.google.com/site/access101site/
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Access full lesson containing this video at: http://www.yourteacher.com/prealgebra/discountformula.php Students solve word problems using the “discount” formula, which states: original price x rate of discount = discount. Students also solve word problems using the “sale price” formula, which states: original price — discount = sale price. For example: A tennis racket marked at $110.50 is on sale at 20% off. What is the discount? What is the sale price? Students also solve word problems using the “rate of discount” formula, which is the same as the percent decrease formula: amount of change/ original number. For example: An old video game is marked down from $48 to $18. What is the rate of discount?
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Access full lesson containing this video at: http://www.yourteacher.com/algebra1/slopeinterceptform.php Students learn to use slope-intercept form to graph a line. Slope-intercept form is y = mx + b form, where m represents the slope, and b represents the y-intercept. So if the equation of a line is y = 3/4 x — 2, then the line is written in y = mx + b form, with m = 3/4 and b = -2. To graph the line, start with the y-intercept, or b, of –2. From there, take the slope, or m, of 3/4, plot a second point, and graph the line.
Duration : 0:2:4
Access full lesson containing this video at: http://www.yourteacher.com/algebra2/pointslopeform.php Students learn to write the equation of a line given a point on the line and the slope of the line, by using the point-slope formula, which states that given a point (x1, y1) and a slope m, the equation of the line can be written as y – y1 = m(x – x1). For example, if a line passes through the point (3, -2), and has a slope of 4, its equation can be written: y + 2 = 4(x – 3). From here, the equation can be converted to either slope-intercept or standard form. Note that a horizontal line passing through a given point has the equation y = the y-coordinate of the given point, and a vertical line passing through a given point has the equation x = the x-coordinate of the given point.
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Access full lesson containing this video at: http://www.yourteacher.com/algebra1/completingthesquare.php Students learn to solve quadratic equations by completing the square. For example, to solve the equation s^2 – 6s + 5 = 0, the first step is put the constant term on the opposite side of the equation as the terms that contain the variables, by subtracting 5 from both sides, to get s^2 – 6s ___ = – 5 ___. Next, the number that goes in each space comes from half the coefficient of the middle term squared, which in this case is half of -6, or -3, squared, which is +9. So a +9 goes in each space, to get s^2 – 6s + 9 = – 5 + 9. The trinomial on the left side of the equation then factors as (s – 3)(s – 3), or (s – 3)^2, and the right side of the equation simplifies to 4, so the problem now reads (s – 3)^2 = 4. Finally, take the square root of both sides to get s – 3 = plus or minus 4, so s – 3 = 4, or s – 3 = -4, and solving each equation from here, s = 7 or -1.
Duration : 0:3:4
Access full lesson containing this video at: http://www.yourteacher.com/algebra2/thequadraticformula.php Students learn to solve quadratic equations in the form ax^2 + bx + c = 0 using the quadratic formula, which states that x = -b plus or minus the square root of b^2 — 4ac over 2a. For example, to solve the equation x^2 — 3x — 8 = 0, since a = 1, b = -3, and c = -8, the quadratic formula states that x = 3 plus or minus the square root of (-3)^2 — 4(1)(-8) over 2(1). Simplifying on the right side of the equation, x = 3 plus or minus the square root of 9 + 32 over 2, or x = 3 plus or minus the square root of 41 over 2, which is the final answer. Note that students also learn to derive the quadratic formula using completing the square in the first example problem in this lesson.
Duration : 0:3:39