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An irrational number cannot be written as a ratio of two integers, and its decimal expansion never terminates or repeats. The square root of 2 has a non-terminating, non-repeating decimal, so it is irrational, while the other three are rational.
When a power is raised to a power, multiply the exponents: (2³)² = 2⁶. Then add exponents when multiplying like bases: 2⁶ · 2⁻⁴ = 2² = 4.
Factor 50 as 25 · 2, then use the product rule for radicals: √50 = √25 · √2 = 5√2. Simplifying a radical means removing the largest perfect-square factor from the radicand.
The complex conjugate of a + bi is a − bi, obtained by changing the sign of the imaginary part only. For 4 − 7i, the real part stays 4 and the imaginary part becomes +7i.
A set is closed under an operation when applying that operation to any two members always gives a member of the set. Subtracting any integer from another integer always yields an integer, but the other sets produce counterexamples such as 3 − 5 = −2.
Scientific notation writes a number as a × 10ⁿ with 1 ≤ a < 10. Moving the decimal point four places right to get 3.6 requires the exponent −4, giving 3.6 × 10⁻⁴.
Expand using the distributive property: 3·1 + 3·(−4i) + 2i·1 + 2i·(−4i) = 3 − 12i + 2i − 8i². Since i² = −1, the last term becomes +8, giving 11 − 10i.
Divide the coefficients and subtract the exponents: (5 ÷ 2) × 10^(6 − (−3)) = 2.5 × 10⁹. Subtracting a negative exponent is equivalent to adding its absolute value.
Raising a power to a power multiplies the exponents: (2/3)·(3/4) = 6/12 = 1/2. So the expression simplifies to x^(1/2), valid because x is positive.
The modulus of a + bi is √(a² + b²). Here a² + b² = 25 + 144 = 169, and √169 = 13.
Multiply numerator and denominator by √3: 6√3/(√3·√3) = 6√3/3 = 2√3. Multiplying by the radical creates a perfect square in the denominator, removing the radical.
Let x = 0.272727…; then 100x = 27.2727…, and subtracting gives 99x = 27, so x = 27/99 = 3/11. A repeating decimal always equals a rational number.
Any integer power of a nonzero rational number is rational, because the reciprocal of a rational is rational and integer powers preserve the ratio of integers. Negative exponents produce reciprocals, which remain rational.
This is a difference of squares: (√7)² − (√3)² = 7 − 3 = 4. The cross terms cancel because they are opposites.
The sum and product of two fractions are always fractions, so the rationals are closed under addition and multiplication. Division is the exception when the divisor is zero, and square roots may be irrational.
Powers of i repeat in a cycle of four: i, −1, −i, 1. Since 43 = 4·10 + 3, i⁴³ = i³ = −i.
Closure fails when the product of two irrational numbers is not irrational. Since √2 is irrational but √2 · √2 = 2 is rational, this pair is a valid counterexample.
Rewrite 8 as 2³ so the equation becomes 2^(x+1) = 2^(3x−3). Equating exponents gives x + 1 = 3x − 3, so 2x = 4 and x = 2.
Compute (1 + i)² = 1 + 2i + i² = 2i, which is not real, so n = 2 fails. Instead, use polar form: 1 + i has argument 45°, and a power is real when n·45° is a multiple of 180°, giving the smallest n = 4.
The number 1/√2 + i/√2 lies on the unit circle with argument 45°. De Moivre's theorem gives argument 8·45° = 360°, so the value is cos 360° + i sin 360° = 1.
When multiplying powers with the same base, add the exponents: x^3 · x^5 = x^8. Dividing by x^2 subtracts the exponent, giving x^(8−2) = x^6.
We need two integers whose product is +12 and whose sum is −7. The pair −3 and −4 satisfies both conditions, so the factorization is (x − 3)(x − 4).
The discriminant is b^2 − 4ac. Here a = 2, b = 3, c = −5, so it equals 9 − 4(2)(−5) = 9 + 40 = 49.
Expand the left side to get 10x − 15 = 4x + 9. Subtract 4x and add 15 to both sides to obtain 6x = 24, so x = 4.
Set 3x − 7 = 5 and solve: 3x = 12, so x = 4. Equivalently, f⁻¹(y) = (y + 7)/3, and (5 + 7)/3 = 4.
The first term is 3 and the common difference is 4, so the 20th term is 3 + 19(4) = 79. The sum is (20/2)(3 + 79) = 10 · 82 = 820.
Factor the numerator as (x − 3)(x + 3) and the denominator as (x + 4)(x − 3). Canceling the common factor (x − 3) leaves (x + 3)/(x + 4).
Vertex form y = a(x − h)^2 + k has vertex (h, k). Here h = −1 and k = 5, so the vertex is (−1, 5), and the negative leading coefficient a = −2 makes the parabola open downward.
Subtract 6 from both sides to get −3x > 9. Dividing by −3 reverses the inequality sign, giving x < −3.
Combine the logarithms to get log₂(x(x − 2)) = 3, so x^2 − 2x = 8, giving x^2 − 2x − 8 = 0. This factors as (x − 4)(x + 2) = 0; x = −2 is rejected because logarithms require positive arguments, leaving x = 4.
Dividing by (x − 2) gives the quotient x^2 − 2x − 3, which factors as (x − 3)(x + 1). Therefore the roots are x = 2, x = 3, and x = −1.
By the Remainder Theorem, the remainder equals p(2). Substituting gives 8 + 8 − 10 + 1 = 7.
The sum of a geometric series is a(r^n − 1)/(r − 1). Here 5(3^4 − 1)/(3 − 1) = 5(80)/2 = 200.
Composition (f ∘ g)(x) means f(g(x)). Substituting gives (2x − 3)^2 + 1 = 4x^2 − 12x + 9 + 1 = 4x^2 − 12x + 10.
A quadratic has exactly one real solution when its discriminant is zero. Here b^2 − 4ac = k^2 − 36 = 0, so k^2 = 36 and k = ±6.
Expand using the distributive property: 8 − 2i + 12i − 3i^2. Since i^2 = −1, this becomes 8 + 10i + 3 = 11 + 10i.
Factor the numerator as (x − 2)(x + 2). For x ≠ 2 the expression simplifies to x + 2, whose limit as x approaches 2 is 4.
From the second equation x = y + 1. Substituting into the first gives 2(y + 1) + 3y = 12, so 5y = 10 and y = 2, hence x = 3.
Exponential decay with half-life 8 means multiplying by 1/2 once per 8 years. The number of half-lives in t years is t/8, so the mass is 40 · (1/2)^(t/8).
Rewrite 8 as 2^3, so the right side becomes 2^(3x−3). Equating exponents gives x + 1 = 3x − 3, so 2x = 4 and x = 2.
Complete the square: x^2 − 6x + 5 = (x − 3)^2 − 4. The squared term is never negative, so the minimum is −4, occurring at x = 3.
The inductive step requires assuming the statement holds for an arbitrary k and then proving it holds for k + 1. Combined with the base case P(1), this establishes the result for all positive integers.
Vertical asymptotes occur where the denominator is zero and the numerator is nonzero, giving factors (x − 2) and (x + 3). A horizontal asymptote at y = 1 requires the numerator and denominator to have equal degree with equal leading coefficients, which the quadratic-over-quadratic form satisfies.
By Vieta's formulas, α + β = 5/2 and αβ = 1/2. Then α^2 + β^2 = (α + β)^2 − 2αβ = 25/4 − 1 = 21/4.
Testing x = 2 gives 8 − 12 + 4 = 0, so (x − 2) is a factor. Dividing yields x^2 − x − 2 = (x − 2)(x + 1), so f(x) = (x − 2)^2(x + 1). Thus x = 2 is a double root and x = −1 is simple, giving exactly 2 distinct real solutions.
For ax^2 + bx + c = 0 the discriminant is b^2 − 4ac. Here a = 3, b = −4, c = 5, so the discriminant equals 16 − 4(3)(5) = 16 − 60 = −44. A negative discriminant confirms the equation has no real roots.
A geometric sequence has nth term a·r^(n−1), where a is the first term and r the common ratio. The first term is 2 and the ratio is 6/2 = 3, so the nth term is 2·3^(n−1). Checking n = 1 gives 2 and n = 2 gives 6, which matches the sequence.
Multiply the coefficients: 3 · (−2) = −6. For each variable add the exponents: x^2 · x^4 = x^6 and y^3 · y^1 = y^4. The product is therefore −6x^6y^4.
Write y = 5x − 2 and solve for x: add 2 to both sides, then divide by 5, giving x = (y + 2)/5. Swapping the variable names gives f⁻¹(x) = (x + 2)/5. Checking, f(f⁻¹(x)) = 5·(x + 2)/5 − 2 = x, as required.
Rewrite the logarithmic equation in exponential form: 2x − 1 = 3^2 = 9. Solving gives 2x = 10, so x = 5, which also satisfies the domain requirement 2x − 1 > 0. Substituting back, log₃(9) = 2 confirms the answer.