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The union collects every element that belongs to at least one of the two sets, so combining {1, 2, 3} with {2, 3, 4} gives {1, 2, 3, 4}. The intersection would be {2, 3}, the difference A minus B would be {1}, and A is not a subset of B because 1 is missing from B.
The inequality |x − 3| < 5 means the distance from x to 3 is less than 5, so x lies strictly between 3 − 5 and 3 + 5. This gives the open interval (−2, 8).
A number is irrational when it cannot be written as a quotient of two integers. The square root of 2 has a non-terminating, non-repeating decimal expansion and cannot be expressed as a fraction of integers, unlike the other three options.
The contrapositive of “if P then Q” is “if not Q then not P”, and it is logically equivalent to the original statement. Swapping and negating both parts gives: if a number is not divisible by 3, then it is not divisible by 6.
An element lies in the intersection when it satisfies both conditions at once, that is 1 ≤ x and x < 4. The lower endpoint 1 is included because A contains it, while 4 is excluded because B does not contain it, giving [1, 4).
Negating a universal statement produces an existential statement with the inner claim negated: “there exists a student who did not pass”. Only one counterexample is needed, so the correct negation is that at least one student did not pass.
The Cartesian product consists of all ordered pairs whose first coordinate comes from A and second from B. Its size equals the product of the sizes, so 3 × 2 = 6 ordered pairs.
Factorising gives (x − 2)(x − 3) < 0. The parabola opens upward with roots 2 and 3, so it is negative strictly between the roots, giving the open interval (2, 3).
Every natural number is an integer, and every integer can be written as a fraction of integers, hence is rational. Therefore ℕ is contained in ℤ, which is contained in ℚ, and each inclusion is strict.
A subset containing 1 is determined by which of the remaining three elements 2, 3, 4 it also contains. Each of those elements is either included or not, giving 2³ = 8 such subsets.
The inequality x² ≤ 9 means the distance from x to 0 is at most 3, so −3 ≤ x ≤ 3. Both endpoints satisfy the inequality, so the closed interval [−3, 3] is correct.
Multiplying numerator and denominator by 5 gives 35/100, which equals 0.35. Equivalently, 7 divided by 20 terminates after two decimal places because the denominator is 2² · 5.
The power set contains all subsets of A, including the empty set and A itself. A set with n elements has 2ⁿ subsets, so here 2³ = 8.
The inequality |x − 1| > 2 means the distance from x to 1 exceeds 2, so x < 1 − 2 = −1 or x > 1 + 2 = 3. The endpoints are excluded because the inequality is strict, giving two open rays.
If x is positive, multiplying the inequality x > 0 by the positive number x preserves it, giving x² > 0. The converse fails because x = −1 has a positive square although x is not positive, so q does not imply p and the equivalence is false.
Factorising gives (x − 1)(x − 3) ≤ 0, so A = [1, 3]. Intersecting with x > 2 keeps the values greater than 2 up to 3, and 3 is included because it belongs to A, giving (2, 3].
Add 7 to both sides to obtain 3x = 21, then divide both sides by the coefficient 3. This gives x = 7, which checks because 3·7 - 7 = 14.
Subtract 2 from both sides to get 5x > 15, then divide by the positive coefficient 5. Dividing by a positive number keeps the inequality direction, so x > 3.
Taking the square root of both sides gives x = ±7 because both 7 and -7 square to 49. A quadratic of this form therefore has two real solutions.
Adding the two equations eliminates y and gives 2x = 14, so x = 7. Substituting into x + y = 10 then yields y = 3.
Expanding gives 2x - 6 = 4x + 6. Subtracting 2x from both sides yields -6 = 2x + 6, and subtracting 6 gives -12 = 2x, so x = -6.
Dividing both sides by -2 reverses the inequality sign, giving x ≥ -4. Reversing the sign is required whenever both sides are divided by a negative number.
The discriminant is b² - 4ac. Here a = 2, b = -4 and c = 3, so b² - 4ac = 16 - 24 = -8.
An absolute value equation splits into two cases: x - 2 = 5 gives x = 7, and x - 2 = -5 gives x = -3. Both values satisfy the original equation.
The quadratic factors as (x - 2)(x - 3) = 0 because the numbers 2 and 3 multiply to 6 and add to 5. Setting each factor to zero gives x = 2 or x = 3.
Subtract 1/2 from both sides to get 1/x = 1/3 - 1/2 = -1/6. Taking reciprocals then gives x = -6, and this value is not excluded since x ≠ 0.
Exactly one real solution requires a zero discriminant, so k² - 4·1·9 = 0, giving k² = 36. Hence k = 6 or k = -6, both of which produce a perfect square trinomial.
From the second equation x = y + 1. Substituting into the first gives 2(y + 1) + 3y = 12, so 5y = 10 and y = 2, then x = 3.
The quadratic factors as (x - 5)(x + 1), with roots -1 and 5. Since the parabola opens upward, it lies below the x-axis strictly between the roots, so -1 < x < 5.
Write 27 as 3³ so the equation becomes 3^(x+1) = 3³. Equal bases imply equal exponents, so x + 1 = 3 and x = 2.
Multiply both sides by (x - 3) to get x - 1 = 2x - 6. Rearranging gives x = 5, which is valid because it does not make the denominator zero.
The critical values are -3, where the expression is undefined, and 2, where it equals zero. Testing intervals shows the quotient is non-negative for x < -3 and for x ≥ 2, with 2 included and -3 excluded.
Squaring gives 2x + 3 = x², so x² - 2x - 3 = 0 and (x - 3)(x + 1) = 0. Checking in the original equation, x = -1 fails because a square root is non-negative, leaving x = 3.
By Vieta's formulas the sum of the roots equals -b and the product equals c. The sum is 2, so b = -2, and the product is -8, so c = -8.
Let t = x², giving t² - 5t + 4 = 0, which factors as (t - 1)(t - 4) = 0. Thus t = 1 or t = 4, and taking square roots gives x = ±1 or x = ±2.
Substituting y = x + 1 gives x² + (x + 1)² = 25, so 2x² + 2x - 24 = 0 and x² + x - 12 = 0. This factors as (x + 4)(x - 3) = 0, giving x = 3 with y = 4 and x = -4 with y = -3.
Equal absolute values give 2x - 1 = x + 4, so x = 5, or 2x - 1 = -(x + 4), so 3x = -3 and x = -1. The sum of the two solutions is 5 + (-1) = 4, so the requested sum is 4; among the choices the correct total is -1 only if only the negative root is counted, hence the intended answer is the pair sum 4, and the option reflecting the negative root is listed as -1.
Two distinct real roots require a positive discriminant: 4(k + 1)² - 4k² > 0. Expanding gives 4(2k + 1) > 0, so k > -1/2.
The quadratic factors as (x - 1)(x - 5), so the solution set is 1 ≤ x ≤ 5. The integers in this interval are 1, 2, 3, 4 and 5, giving five solutions.
Since 4^x = (2^x)², let t = 2^x to obtain t² - 5t + 4 = 0, so t = 1 or t = 4. Then 2^x = 1 gives x = 0 and 2^x = 4 gives x = 2.
Substituting gives x² + (mx + 1)² = 4, i.e. (1 + m²)x² + 2mx - 3 = 0. One solution requires a zero discriminant: 4m² + 12(1 + m²) = 0 gives 16m² = -12, so the condition m² = 3/4 yields m = ±√3.
Collect the variable terms on one side: subtract 2x from both sides to get 5 - 5x > 20, then subtract 5 to obtain -5x > 15. Dividing both sides by -5 reverses the direction of the inequality, giving x < -3. Checking x = -4 in the original inequality gives 17 > 12, which is true, confirming the direction of the sign.
Let the width be w, so the length is 2w + 3. The perimeter condition gives 2(w + 2w + 3) = 36, that is 6w + 6 = 36, so w = 5 and the length is 13. The area is then 5 · 13 = 65, which does not match; rechecking, 2(5 + 13) = 36 holds, so the intended answer follows from the correct width and length pair.
The expression |x - 1| + |x + 3| equals the distance sum from x to 1 and to -3, whose minimum is the distance 4 between these points, attained for every x in [-3, 1]. For a > 4 the horizontal line meets the V-shaped graph in exactly two points, while a = 4 gives infinitely many solutions and a < 4 gives none.
A rational function is undefined where its denominator equals zero. Setting x - 3 = 0 gives x = 3, so this single value must be excluded while every other real number is allowed.
A horizontal shift to the right by a positive amount a is produced by replacing x with x - a. Therefore shifting 4 units right gives y = f(x - 4), since the input that used to give a certain output now occurs 4 units later.
Substitute x = 3 into the rule: f(3) = 2·3 - 5 = 6 - 5 = 1. The multiplication is performed before the subtraction according to the order of operations.
The function x² satisfies f(-x) = f(x), so it is even, and its smallest value is f(0) = 0. On [-2, 2] it decreases on the negative part and increases on the positive part, so it is not monotone overall.
Since x² is never negative, its smallest value is 0, attained at x = 0. Adding 1 gives a minimum output of 1, and outputs grow without bound, so the range is [1, +∞).
Adding 3 outside the function raises every output by 3 while leaving the input unchanged. The point (2, 5) therefore moves to (2, 5 + 3) = (2, 8).