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For Exercises 41–54, write the equation in the form
. Then if the equation represents a
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ALEKS 360 COLLLEGE ALGEBRA ACCESS
- For Exercises 109–111, (a) evaluate the discriminant and (b) determine the number and type of solutions to each equation. 109. 4x – 20x + 25 = 0 110. -2y = 5y – 1 111. 5t(t + 1) = 4t – 11arrow_forwardFor Exercises 7–12, write an equivalent equation using pola coordinates. 7. x + 5y = 12 9. x + (y – 3)? = 9 8. x = 12 10. x + y = 100 %3D 11. y = -4 12. y = xarrow_forward–4(–6m – 7).arrow_forward
- Rewrite y in the more convenient form, y = + C.arrow_forward2. Expand and simplify to express each equation in standard form. [2, 2, 2, 3, 3, 2] a) y = (x + 5)(x + 2) b) y = (x- 4)(x+ 3) c) y = (2x- 3)(x-6) d) y = (3x + 4)2 e) y = 3(x – 4)2 +1 f) y = (4x + 3)(4x- 3)arrow_forwardDetermine when y = In(2x² – 4x).arrow_forward
- 4. Complete the square to write y = -2 x2 + 16 x – 21 in the form y = a (x - h)² + k %3D %3Darrow_forwardGive your answers in the form ax + by = c. 1 a AB: y = 2x+ 11; midpoint (-4, 3) b AB: y =-3x + 7; midpoint (6, -11) 2 a AB : y=-x-10; midpoint (-12, 6) 3 b AB : y =x+1; midpoint (10, 5) b AB: 3x – 4y = 9; midpoint (-1, -3) 9 7 4' 2 3. a AB: 4x + 5y = 13; midpoint (2, 1) 4 a AB: 6x – 24y =-37; midpoint b AB: 4x + 12y = 33; midpoint For questions 5 to 8, use the method demonstrated in Worked Example 14.5 to find the equation of the perpendicular bisector of the line segment connecting the given points. Give your answers in the form y=mx +c. 5 a (5, 2) and (13, 6) b (3, 1) and (21, 7) 6 a (-4, 5) and (5, -1) b (-2, -3) and (2, 7) (17 1) a and 21 b 4 and (9 34 7 (2'5, 23 31 and 7 17 (11 13 and 8. a 9 For each of the following Voronoi diagrams, give all the examples of the: i sites iii finite edges ii vertices iv finite cells.arrow_forwardThe function, f(x) = –2x² + x + 5, is in standard form. The What are the zeros of the function quadratic equation is 0 = -2x + x + 5, where a = -2, b = 1, and c = 5. The discriminate b2 – 4ac is 41. Now, complete step 5 to solve for the zeros of the quadratic function. f(x) = x + 5 – 2x2? x = -1±/4T -4 5. Solve using the quadratic formula. O x= 1±/41 -4 -b±yb²- 4ac_ x = 2a x = -1±/39 - 4 X = 1±/39 - 4arrow_forward
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