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-20x 5y equals 40 changed to slope-intercept form?
Oh, dude, you just gotta rearrange that equation a bit. So, like, first divide by 5 to get -4x - y = 8. Then, if you wanna be all fancy and use slope-intercept form, just solve for y to get y = -4x + 8. And there you have it, a technically correct answer with a sprinkle of my signature humor.
How do you solve 20x plus 5y equals 15?
If: 20x+5y = 15 Then: y = -4x+3
What does x equal when x plus 20x equals to 60000?
x+20x = 60000 21x = 60000 x = 60000/21 = 2857.'142857' recurring
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
Which is the general form of the equation 5x 2 - 20x equals 15?
I dont see any related links or files that I need to follow as indicated by your question. the general form of 5x2 - 20x = 15 is... 5x2 - 20x - 15 = 0 ax2 + bx + c = 0
