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An irrational number cannot be written as a ratio of two integers and has a non-terminating, non-repeating decimal expansion. The square root of 2 is not a perfect square, so its decimal expansion never terminates or repeats, making it irrational.
The integers are the whole numbers together with their additive inverses, so they include negatives, zero, and positives but exclude fractions and decimals. Counting numbers omit zero and negatives, and fractions belong to the rationals rather than the integers.
Absolute value measures distance from zero on the number line, so |−7| = 7 and |3| = 3. Adding these non-negative results gives 7 + 3 = 10, which is positive regardless of the sign inside the bars.
The additive inverse of a number is the value that sums with it to zero. Since −2/5 + 2/5 = 0, the additive inverse is 2/5, obtained by changing the sign only.
π has a decimal expansion that never terminates and never repeats, so it cannot be expressed as a ratio of integers. The other choices all equal exact rational values.
Factor 72 as 36 · 2, and since 36 is a perfect square, √72 = √36 · √2 = 6√2. The largest perfect-square factor must be extracted to reach simplest form.
Convert to decimals: −3/4 = −0.75 and √0.25 = 0.5. On the number line −0.9 is farther left than −0.75, and 0.2 is less than 0.5, giving the order −0.9, −0.75, 0.2, 0.5.
A fraction in lowest terms terminates exactly when its denominator factors into 2s and 5s, because those primes produce powers of 10. A denominator such as 3 or 7 introduces repeating digits instead.
Multiply coefficients and radicands separately: 2 · 5 = 10 and √3 · √12 = √36 = 6. Then 10 · 6 = 60, so the product is the integer 60.
If a rational plus an irrational gave a rational result, subtracting the rational would make the irrational equal to a rational difference, a contradiction. Therefore the sum of a rational and an irrational is always irrational.
Evaluate each absolute value first: |−8| = 8, |−3| = 3, and |5| = 5. Then 8 − 3 + 5 = 10, following left-to-right order of operations.
√9 equals exactly 3, which is an integer and therefore also rational since 3 = 3/1. The other choices are irrational and cannot be written as ratios of integers.
The side length is √50. Since 50 = 25 · 2 and 25 is a perfect square, √50 = √25 · √2 = 5√2 units.
The rationals are dense, so the average of two distinct rationals is another rational strictly between them. Repeating this process shows infinitely many rationals lie in any interval.
This is a difference of squares: (√5)² − (√2)² = 5 − 2 = 3. The cross terms +√10 and −√10 cancel, leaving a rational result.
Let x = 0.272727…. Then 100x = 27.272727…, and subtracting gives 99x = 27, so x = 27/99 = 3/11 in lowest terms. The repeating block length of 2 produces the denominator 99.
The leading coefficient is the coefficient of the term with the highest power of the variable. Here the highest power is x³, whose coefficient is 7. The constant −9 and the coefficient −4 belong to lower-degree terms and do not qualify.
For a polynomial in several variables, the degree of a term is the sum of the exponents in that term, and the degree of the polynomial is the largest such sum. The term 5x²y³ has degree 2 + 3 = 5, which exceeds the degree 2 of −2xy. Hence the polynomial has degree 5.
Solving for y gives 2y = −4x + 10, so y = −2x + 5. In slope-intercept form y = mx + b, the coefficient m is the slope, which is −2. The coefficient 4 belongs to x before rearrangement and is not the slope.
A pair solves the equation only if substituting both coordinates makes the two sides equal. For (3, 2), we get 3(3) − 2 = 9 − 2 = 7, which matches the right side. Checking the other pairs gives values other than 7.
The equation y = 5 fixes the y-coordinate at 5 for every value of x, so all points have the form (x, 5). Such points lie on a horizontal line crossing the y-axis at (0, 5). Its slope is 0 because y does not change as x changes.
When adding polynomials, combine like terms by adding coefficients of matching powers. The x² terms give 2 + 1 = 3, the x terms give −3 + 5 = 2, and the constants give 1 − 4 = −3. The sum is therefore 3x² + 2x − 3.
The y-intercept is the point where the line crosses the y-axis, so set x = 0. Then −3y = 12, giving y = −4, and the intercept is (0, −4). Substituting any other listed value fails to satisfy the equation at x = 0.
A binomial has exactly two terms, and its degree is the highest exponent appearing. The expression x³ − 2x has two terms and highest power 3, so it is a binomial of degree 3. The other choices have either three terms or a different degree.
Applying the distributive property, x(x − 3) + 4(x − 3) = x² − 3x + 4x − 12. Combining the like x terms gives x² + x − 12. The constant is the product 4 · (−3) = −12.
The slope formula is m = (y₂ − y₁)/(x₂ − x₁). Substituting gives (−1 − 5)/(4 − (−2)) = −6/6 = −1. A negative slope matches the fact that y decreases as x increases between these points.
Point-slope form is y − y₁ = m(x − x₁). With m = 4 and the point (3, −2), substitute y₁ = −2 to get y − (−2) = 4(x − 3), which simplifies to y + 2 = 4(x − 3). The signs of both coordinates must be substituted carefully.
Adding the two equations eliminates y: (2x + y) + (x − y) = 11 + 1, so 3x = 12 and x = 4. Substituting back gives y = 3, confirming the pair (4, 3) satisfies both equations.
Rewrite both in slope-intercept form: the first becomes y = −½x + 2, and the second becomes y = −½x + 2.5. The slopes are equal but the y-intercepts differ, so the lines are parallel and never meet. Equal slopes with different intercepts mean no solution.
We need two numbers whose product is 6 and whose sum is −5. The numbers −2 and −3 satisfy both conditions, so x² − 5x + 6 = (x − 2)(x − 3). Expanding confirms the middle term −5x and constant +6.
A perpendicular line has slope equal to the negative reciprocal of 2, which is −½. Using point-slope form with (4, 1): y − 1 = −½(x − 4), which simplifies to y = −½x + 3. Substituting x = 4 gives y = 1, confirming the point lies on the line.
Squaring a binomial follows (a − b)² = a² − 2ab + b². With a = 3x and b = 2, this gives 9x² − 2(3x)(2) + 4 = 9x² − 12x + 4. The middle term is essential and cannot be omitted.
The intercepts give the points (6, 0) and (0, −3). The slope is (0 − (−3))/(6 − 0) = 3/6 = ½, and the y-intercept is −3, so y = ½x − 3. Checking x = 6 gives y = 0, as required.
Dividing the first equation by 2 gives 2x − y = 4, and dividing the second by 3 gives 2x − y = 5. The left sides are identical but the constants differ, so the lines are parallel and distinct. Parallel distinct lines never intersect, so the system has no solution.
Divide each term of the polynomial by 3x: 6x³/(3x) = 2x², −9x²/(3x) = −3x, and 3x/(3x) = 1. The quotient is 2x² − 3x + 1. Each term's exponents decrease by one and coefficients divide by 3.
The point (0, 3) gives c = 3. Substituting (1, 4) gives a + b + 3 = 4, so a + b = 1; substituting (2, 9) gives 4a + 2b + 3 = 9, so 2a + b = 3. Subtracting yields a = 2 and then b = −1, so a = 2.
No solution occurs when the lines are parallel but distinct, meaning the ratios of x-coefficients and y-coefficients are equal while the constant ratio differs. Setting k/2 = 3/1 gives k = 6; then the first equation is 6x + 3y = 6, or 2x + y = 2, which is parallel to 2x + y = 4. The constants differ, so there is no solution.
Dividing x³ − 4x² + x + 6 by (x − 2) gives the quadratic x² − 2x − 3, which factors as (x − 3)(x + 1) because −3 and 1 multiply to −3 and sum to −2. The full factorization is therefore (x − 2)(x − 3)(x + 1). Expanding confirms the original cubic.
The slope is (11 − 5)/(3 − 1) = 3, and using point-slope form y − 5 = 3(x − 1) gives y = 3x + 2. Rearranging to standard form yields 3x − y = −2, where A = 3 is positive and all coefficients are integers.
Use the identity x² + y² = (x + y)² − 2xy. Substituting gives 10² − 2(21) = 100 − 42 = 58. The values x = 7 and y = 3 satisfy both conditions and give 49 + 9 = 58, confirming the result.
Setting the expressions equal gives x² + (2 − m)x − 3 = 0. A single intersection means this quadratic has a zero discriminant: (2 − m)² − 4(1)(−3) = 0, so (2 − m)² = −12, which has no real solution. Rechecking by setting x² + 2x + 1 = mx + 4 gives x² + (2 − m)x − 3 = 0, and for m = 0 the discriminant is 4 + 12 = 16, so the line y = 4 meets the parabola twice, not once; among the listed values none yields exactly one intersection.
Two distinct nonparallel lines in the plane always intersect in exactly one point, and lines with different slopes are never parallel. A linear system's solution set is the intersection of its graphs, so the single intersection point gives exactly one solution. Systems with no solution or infinitely many solutions arise only when the slopes are equal.
For a monic quadratic x² + bx + c = 0, the roots sum to −b and multiply to c. Here the sum is 7 and the product is 12, and 3 and 4 satisfy both conditions.
The common difference of an arithmetic progression is found by subtracting any term from the next term. Here 11 − 5 = 6, and the same difference holds for 17 − 11 and 23 − 17.
The discriminant of ax² + bx + c = 0 is b² − 4ac. Substituting a = 2, b = 5, c = −3 gives 25 − 4(2)(−3) = 25 + 24 = 49.
Using aₙ = a + (n − 1)d with n = 10 gives 47 = 2 + 9d, so 9d = 45 and d = 5. Substituting back confirms the tenth term.
The discriminant is 16 − 16 = 0, so the equation has a repeated real root. Factorising gives (x + 2)² = 0, giving x = −2 twice.
Rearranging gives x² = 9, and taking the square root of both sides yields x = ±3. Both values satisfy the original equation.
The first term is found by substituting n = 1 into the formula, giving 3(1) + 4 = 7. The constant 4 alone is not the first term.
Substituting n = 5 into Sₙ = n² + 2n gives 25 + 10 = 35. The formula already encodes the arithmetic structure.