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How do you solve x 2-18x-4 equals 0 by completing the square?
x² - 18x - 4 = o x² - 18x = 4 x² - 18x + (18/2)² = 4 + (18/2)² x² - 18x + 81 = 4 + 81 (x - 9)² = 85 x - 9 = ±√85 x = 9 ±√85 x = 9 + √85 or x = 9 - √85 if you are still confused, i want you to follow the related link that explains the concept of completing the square clearly.
What is a algebra puzzle?
An equation to figure out what the answer of "x" is. eg, Removing the brackets - 6(3x-2) = 2(8x+4) 18x-12 = 16x + 8 18x-16x = 8+12 2x = 20 x = 10
Can you Factorise 20x-28?
Yes and it is 4(5x-7) when factored
Find the area S of the region bounded by the parabola y 5x2 the tangent line to this parabola at 2 20 and the x-axis?
First we need to find the equation of the tangent line to the parabola at (2, 20).Step 1. Take the derivative of the function of the parabola.Let f(x) = 5x^2f'(x) = 10xStep 2. Find the slope of the tangent line at x = 2. Evaluate f'(2).f'(2) = 2 x 10 = 20Step 3. Using the slope, m = 20, and the point (2, 20), find the equation of the tangent line at that point. Use the point-slope form of a line(y - y1) = m(x - x1)(y - 20) = 20(x - 2)y - 20 = 20x - 40 add 20 to both sidesy = 20x - 20Step 4. Find the points of intersections of y = 5x^2 and y = 20x - 205x^2 = 20x - 20 Divide by 5 to both sidesx^2 = 4x - 4 subtract 4x and add 4 to both sidesx^2 - 4x + 4 = 0 factor(x - 2)^2= 0x = 2Step 5. Find the intersection of the tangent line with x-axis.y = 20x - 20y = 020x - 20 = 0x = 1Since the vertex of the parabola is (0, 0) and the intersection of the tangent line with parabola is (2,20) we use the interval [0, 2] to fin the required area.Step 6. IntegrateA = ∫ [(5x^2)] dx, where the below boundary is 0, and the upper boundary is 2 minus A= ∫ (20x + 20)] dx from 1 to 2= 10/3
The perimeter around a dog's running space in 20x 2 28x 8 The length of the dog's running space is 10x 4 what is the width of the dog's running space Show why your answer is correct?
The question is utter nonsense! The perimeter of is a 1-dimensional measure so it cannot be 20x 2 28x 8. Similarly, the length of the running space is also a linear measure and so it cannot be 10x 4. There is, therefore, no way that the answer can be correct.
