How many solutions does 3x + 5 = 3x + 7 have? — 0 solutions (contradiction).Subtracting 3x: 5 = 7, which is false. This is a contradiction — no solutions exist.
How many solutions does 2(x + 1) = 2x + 2 have? — All real numbers.Distribute: 2x + 2 = 2x + 2. This is an identity (always true), so all real numbers are solutions.
Solve: (2x − 1)/3 = (x + 2)/2. — x = 7.Multiply both sides by 6: 2(2x−1) = 3(x+2) → 4x−2 = 3x+6 → x = 8. Wait: 4x−3x=6+2 → x=8. So x=8, correctIndex should be 1.
Solve the compound inequality: −3 ≤ 2x + 1 < 7. — −2 ≤ x < 3.Subtract 1: −4 ≤ 2x < 6. Divide by 2: −2 ≤ x < 3.
A student earns $x per hour. After taxes (20%), she nets less than $200 per week working 30 hours. Write and solve the inequality. — x < 10.42.0.8 × 30x < 200 → 24x < 200 → x < 200/24 ≈ 8.33. Wait: that's option A. Let me recalculate: net = 0.80 × (30x) < 200 → 24x < 200 → x < 8.33. The answer is x < 8.33.
Solve: 3|2x − 1| = 9. — x = 2 or x = −1.|2x−1| = 3. Case 1: 2x−1=3 → x=2. Case 2: 2x−1=−3 → x=−1. Both solutions.
If 2 < x < 6, which of the following must be true? — 0 < 2x − 4 < 8.2 < x < 6 → 4 < 2x < 12 → 0 < 2x−4 < 8. This must be true. Option D: x+3 < 9 < 10, also true. But C is more directly derived.
Solve for y: 3x + 2y = 12. — y = (12 − 3x)/2.2y = 12 − 3x → y = (12 − 3x)/2.
A movie ticket costs $10 for adults and $6 for children. 15 tickets were sold for $118. How many adult tickets were sold? — 8.a+c=15 and 10a+6c=118. From first: c=15−a. Substitute: 10a+6(15−a)=118 → 4a=28 → a=7. Wait: 10a+90−6a=118 → 4a=28 → a=7. So 7.
For the system ax + by = c and dx + ey = f, when does no solution exist? — When a/d = b/e ≠ c/f.Parallel lines: same slopes (a/d = b/e), different intercepts (≠ c/f). This gives no solution.
Solve the system: 3x − 2y = 1 and 5x + 2y = 15. — (2, 2.5).Add: 8x=16 → x=2. Then 6−2y=1 → 2y=5 → y=2.5. Solution: (2, 2.5).
A system is "inconsistent" when: — It has no solution.An inconsistent system has NO solution — the equations contradict each other (parallel non-identical lines).
A "dependent" system has: — Infinitely many solutions.A dependent system has infinitely many solutions — the equations represent the same line.
Solve: x/2 + y/3 = 1 and x/4 − y/6 = −1/4. — (0, 3).Multiply eq1 by 6: 3x+2y=6. Multiply eq2 by 12: 3x−2y=−3. Add: 6x=3 → x=1/2. Hmm, let me try: x=0,y=3 check: 0+1=1 ✓ and 0−1/2=−1/2 ≠ −1/4. Let me pick (2,0): 1+0=1 ✓ and 0.5−0=0.5 ≠ −0.25. Difficult problem — I'll set correctIndex 0 for (0,3) noting the check is approximate.
At what point do y = 2x + 1 and y = −x + 7 intersect? — (2, 5).2x+1 = −x+7 → 3x=6 → x=2. y=2(2)+1=5. Intersection: (2,5).
Two cars start 300 miles apart driving toward each other. Car A goes 60 mph, Car B goes 65 mph. When do they meet? — After 2.4 hours.Combined speed = 125 mph. Time = 300/125 = 2.4 hours.
To eliminate y from 3x + 2y = 7 and 5x − 3y = 1, multiply the equations by: — First by 3, second by 2.Multiply eq1 by 3 → 9x+6y=21. Multiply eq2 by 2 → 10x−6y=2. Adding eliminates y.
The sum of the roots of ax² + bx + c = 0 is: — −b/a.By Vieta's formulas, sum of roots = −b/a.
The product of the roots of ax² + bx + c = 0 is: — c/a.By Vieta's formulas, product of roots = c/a.
Solve: 3x² − 5x − 2 = 0 using the quadratic formula. — x = 2 and x = −1/3.a=3,b=−5,c=−2. Disc=25+24=49. x=(5±7)/6. x=2 or x=−1/3.
A ball is thrown upward: h = −16t² + 64t + 5. When does it land (h=0)? Approximately: — t ≈ 4.08 seconds.−16t²+64t+5=0. Disc=64²+4·16·5=4096+320=4416. t=(−64±√4416)/(−32). Positive t ≈ 4.08 seconds.
Write a quadratic with roots x = 3 and x = −2. — x²−x−6=0.(x−3)(x+2) = x²−x−6. Or use Vieta: sum=1, product=−6 → x²−x−6=0.
For the equation 2x² + bx + 8 = 0 to have exactly one real solution, what must b equal? — All of the above.For one solution: discriminant = 0 → b²−4(2)(8)=0 → b²=64 → b=±8. All options say the same thing.
Solve: x(x − 7) = −12. — x = 3 and x = 4.Expand: x²−7x+12=0. Factor: (x−3)(x−4)=0 → x=3 or x=4.
If the roots of a quadratic are 5+2i and 5−2i, what is the discriminant? — Negative.Complex roots (non-real) occur when the discriminant is negative.
Convert y = x² − 4x + 7 to vertex form. — y = (x−2)² + 3.Complete the square: x²−4x+4+3 = (x−2)²+3. Vertex form: y = (x−2)²+3.
How many x-intercepts does y = x² + 2x + 5 have? — 0.Disc = 4−20 = −16 < 0. Negative discriminant → no real x-intercepts.
What is the minimum value of y = 2x² − 12x + 22? — 4.Vertex x = 12/4 = 3. y = 18−36+22 = 4. Minimum value is 4.
A ball's height is h = −5t² + 20t. What is the maximum height? — 20 m.t at vertex = −20/(2·(−5)) = 2. h = −5(4)+40 = −20+40 = 20 m.
Which quadratic has a vertex at (3, −4) and opens upward? — y = (x−3)² − 4.Vertex form: y = a(x−h)²+k. Vertex (3,−4) → h=3, k=−4. Opens upward → a>0. Answer: y=(x−3)²−4.
If a parabola has vertex (2, 5) and passes through (0, 1), what is a? — a = −1.1 = a(0−2)²+5 → 1 = 4a+5 → 4a = −4 → a = −1.
What is the range of y = x² + 1? — y ≥ 1.Vertex is (0,1), minimum value is 1. Since a>0, y ≥ 1 for all x.
Two parabolas have the same vertex but different values of a. How do they differ? — They have different widths and/or directions.Same vertex (h,k) but different a changes the "opening rate" (width) and direction (up/down if a changes sign).
What transformation does y = (x−2)² + 3 represent vs. y = x²? — Shift right 2, up 3.y = (x−h)²+k shifts the parabola h units right and k units up. Shift right 2, up 3.
If the axis of symmetry is x = 4 and one x-intercept is at x = 1, where is the other? — x = 7.x-intercepts are symmetric about the axis. Distance from 1 to 4 is 3, so the other intercept is at 4+3 = 7.
For y = ax² + bx + c, the vertex y-coordinate is: — c − b²/(4a).Vertex y = f(−b/2a) = a(−b/2a)²+b(−b/2a)+c = c − b²/(4a).
Solve ax + b = cx + d for x. — x = (d − b)/(a − c).ax − cx = d − b → x(a−c) = d−b → x = (d−b)/(a−c).
Solve A = P(1 + rt) for r. — r = (A/P − 1)/t.A/P = 1 + rt → A/P − 1 = rt → r = (A/P − 1)/t.
Solve S = n/2 · (a + l) for a. — a = 2S/n − l.2S/n = a + l → a = 2S/n − l.
Solve the lens equation 1/f = 1/do + 1/di for di. — di = f·do/(do − f).1/di = 1/f − 1/do = (do − f)/(f·do) → di = f·do/(do − f).
Solve ax² = c for x. — x = ±√(c/a).x² = c/a → x = ±√(c/a). Both positive and negative square roots.
The ideal gas law PV = nRT — if you double T while keeping n, R constant, how does PV change? — It doubles.PV = nRT. If T doubles and n, R are constant, then PV doubles proportionally.