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DSPM 0850 Final Exam Solutions Systems: Ch. 4
The 2 graphs intersect at the point (-1, -3); this is the solution to the system.
5
5
2 x ( x 3.5) 3
6 x (^6) The intersection point is (-0.5, 1)
x y x y
10 11 12 13 14 15 16 17 18 19 20
1 2 3 4 5 6 7 8 9 10 10
0
25 x x 11
10 x 20 The graphs intersect at (18, 7). This is the solution
The solution is (
x y x y x y y y y y x
y y
(^2 ) (^2 5 )
x y x y x x y โ
^ +^ =
5
5
x 1 4 3 x
5 x 5 The shaded region is the solution region.
- stairclimber = x = 8 min; bike = y = 22 minutes;
set up the system as shown below and solve for x & y.
30 11.5 9 290
x y x y
^ +^ =
- Let x = the number of $8 tickets; let y = the number of $ tickets. Solve the following system using either of the three methods. 480 8 12 4620
x y x y
^ +^ =
x = 285; y = 195
Factoring: Ch. 5
- GCF = 5x; 5x(5x โ 6) 45. GCF = 5y; 5y(1 + 3y)
- We need two factors of 12 whose sum is 8 โฆ
(x + 2)(x + 6)
- We need two factors of 50 whose difference is 5; highest number must be negative. (x โ 10)(x + 5)
- Multiply 1st^ term by last term โฆ -54x^2 โฆ we need 2 factors of 54x^2 whose difference is 25 โฆ (+27 & -2) โฆ
9x^2 โ 2x + 27x - 6 โก (9x^2 โ 2x) + (27x - 6)
Factor by grouping โฆ x(9x โ 2) + 3(9x โ 2) โก (x+3)(9x-2)
- Multiply 1st^ and last terms (40x2) we need 2 factors of 40 whose sum is โ22; -20 & -2. Replace the middle term with these factors โฆ 4x^2 โ 20x โ 2x + 10 โก (4x^2 โ20x) + ( โ2x + 10) Factor out GCF: 4x(x โ 5) + -2(x โ 5) โก (4x โ 2)(x โ 5) โก 2(x โ 1)(x โ 5)
- A difference of squares โฆ ( t - 7)( t + 7 ) 62. A difference of squares โฆ (2y โ 3x)(2y + 3x)
- A perfect square trinomial โฆ (x + 2)^2
Quadratics: Ch. 6
- x = -b/2a = 4/2 = 2
y = f(2) = (2)^2 โ 4(2) โ 2 = - vertex is: (2, -6)
- x = -b/2a = -2/2 = - y = f(-1) = 2 + 2(-1) + (-1)^2 = 1 vertex is: (-1, 1)
- Wider (because ยฝ is smaller than 1, the coefficient of x^2 ) โฆ shifted 1 unit to the left and 2 units up
- Narrower (because 2 is greater than 1, the coefficient of x^2 ) โฆ shifted 1 unit to the right and 3 units down
- The LSE factors: (x + 5)(x โ 4) = 0; Use the ZPP to solve โฆ x = {-5, 4}
- The LSE factors: (5x + 2)(3x โ 2) = 0; Use the ZPP to solve โฆ. X = {-2/5, 2/3}
- Rewrite the eqn: 5x^2 โ 5x + 1 = 0; a = 5, b = -5, c = 1; plug these values into the Quad Formula yields:
x = (5 ยฑ sqrt(5))/
- The inequation may be solved by factoring โก (x + 3)(x + 1) = 0 and x = -3 and โ1. These are the roots โฆ Graphing the function we see that the graph is below 0 between these two roots, so: -3 โค x โค -
- The function will solve by factoring โฆ we find the roots are 1/6 and 2. Graph the function and we see that the graph is greater than (above) 0 outside these values, so: x < 1/6 & x > 2
- b. Graph the function in your calculator using the specified window. Enter a second eqn, Y2 = 32. Use the CALC (2nd^ TRACE) and calculate the INTERSECT of the 2 graphs. (x = 1 and 1.75) c. CALC the MAXIMUM โฆ we find that x (time) = 1.375 secs. When the stone is at its maximum height (y) of 34.25 ft.
Undo the radical by squaring both sides and solving for x.
Use Pythagorean Theorem: a^2 + b^2 = c^2 Use Pyth. Th.: 5^2 + b^2 = 8^2 ; solve for b.
