Math II
Connect real and complex numbers, polynomials, quadratics, proof, circles, trigonometry, probability, and modeling.
- Problem types
- 786
- Practice variants
- 3,144
Page 17 of 22
Each problem type has four distinct practice variants. Open a preview to move among all four.
Solve navigation with a signed vector, distance, and bearing
Use trig ratios and the Pythagorean Theorem to solve right triangles in applied problems.
A navigation result combines vector magnitude with a direction measured from a named axis. We’ll encode east and north as perpendicular components, use the Pythagorean theorem for distance, form north-over-east …
Preview problemChoose the correct method for a right-triangle problem: trig ratio, inverse trig, Pythagorean Theorem, or insufficient information
Use trig ratios and the Pythagorean Theorem to solve right triangles in applied problems.
Method choice should follow the kinds of givens and target, not just the fact that the triangle is right. We’ll inventory two known sides and one side target, identify the …
Preview problemRound a right-triangle result to the requested precision
Use trig ratios and the Pythagorean Theorem to solve right triangles in applied problems.
Rounding begins with the requested place, then uses only the digit immediately to its right to decide the change. We’ll locate the tenths digit, inspect the hundredths digit, update the …
Preview problemComplete a closed contextual quantity-and-unit interpretation
Use trig ratios and the Pythagorean Theorem to solve right triangles in applied problems.
A complete contextual conclusion must interpret what the variable measures, not merely repeat its equation. We’ll reconnect the symbol to the named object and vertical quantity, distinguish a length from …
Preview problemFind separate leg and hypotenuse fields in a45-45-90 triangle
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
A forty-five-forty-five-ninety triangle scales the ordered pattern leg, leg, hypotenuse from one, one, square root of two. We’ll use the equal-leg property, scale all three ratio terms by the given …
Preview problemUse the symbolic1:√3:2 relationship in a30-60-90 triangle
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
The thirty-sixty-ninety ratio is ordered by the sides opposite thirty, sixty, and ninety degrees. We’ll map those to short leg, long leg, and hypotenuse, scale the one-to-square-root-of-three-to-two pattern by the …
Preview problemFind missing side lengths in a 45-45-90 triangle using \(x, x, x sqrt(2)\)
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
Equal acute angles make the two legs equal, while the radical factor belongs only to the hypotenuse. We’ll transfer the known leg length to the other leg, scale it by …
Preview problemFind numeric sides in a30-60-90 triangle
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
Starting from the short leg fixes the scale for the entire thirty-sixty-ninety pattern. We’ll align short leg, long leg, and hypotenuse with one, square root of three, and two, multiply …
Preview problemDerive separate exact sine, cosine, and tangent values at45°
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
A unit forty-five-forty-five-ninety triangle supplies all three exact trig values from the same side roles. We’ll form opposite-over-hypotenuse, adjacent-over-hypotenuse, and opposite-over-adjacent, rationalize any radical denominator, and report the simplified values …
Preview problemDerive six separately labeled exact trig values at30° and60°
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
The same thirty-sixty-ninety triangle yields two sets of trig values because opposite and adjacent swap when the reference angle changes. We’ll derive sine, cosine, and tangent for the first angle, …
Preview problemApply the correct special right-triangle family in context
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
The ramp angle identifies the vertical rise as the side opposite thirty degrees, so it is the short leg. We’ll match the context to the thirty-sixty-ninety family, scale the exact …
Preview problemChoose whether to use a 45-45-90 triangle or a 30-60-90 triangle
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
A right triangle’s two acute angles sum to ninety degrees, which identifies its special family. We’ll subtract the known acute angle from ninety, assemble the complete angle set, name the …
Preview problemCompare exact trigonometric values from special right triangles
Derive and use trig ratios for 30-60-90 and 45-45-90 special right triangles.
The comparison must use evaluated trig ratios rather than the angle measures themselves. We’ll recall each exact sine value from its special triangle, place them over a common positive denominator, …
Preview problemSimplify powers of $i$ using the repeating cycle $i, -1, -i, 1$
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Powers of the imaginary unit repeat in a four-step cycle, so the exponent’s remainder locates the value. We’ll build the first four powers, separate complete groups of four from the …
Preview problemRewrite square roots of negative numbers in terms of i
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
A negative radicand can be separated into a positive magnitude and the defining factor for the imaginary unit. We’ll factor out negative one, evaluate the ordinary principal square root, replace …
Preview problemIdentify separate real and imaginary numeric parts
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
In standard form, the real part is the term without the imaginary unit and the imaginary coordinate is the real coefficient multiplying it. We’ll align the expression with a plus …
Preview problemRewrite a simple complex expression in standard form \(a+bi\)
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Standard complex form places the real term first and the imaginary term second without changing either value. We’ll identify the two term types, use commutativity to reorder them, preserve the …
Preview problemClassify a complex number as real, pure imaginary, or complex non-real
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Classification becomes unambiguous after both coefficients are exposed in standard form. We’ll write the number with an explicit zero imaginary coefficient, compare the coefficient pattern with the mutually exclusive category …
Preview problemPlot a complex number \(a+bi\) on the complex plane
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
The complex plane maps a plus b times i to the real ordered pair a comma b. We’ll extract the real part for horizontal motion, use the imaginary coefficient—not the …
Preview problemSolve equations of the form \(x^2 =\) a negative number using \(i\)
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Solving a square equation requires both square-root branches, even when the radicand is negative. We’ll take plus and minus the square root, rewrite the negative root using the imaginary unit, …
Preview problemSimplify expressions with i and write the result in standard complex form
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Real terms and imaginary terms form separate like-term groups. We’ll preserve the real component, combine only the coefficients attached to the imaginary unit, evaluate that signed coefficient, and place the …
Preview problemDecide whether two complex expressions are equal by simplifying to a+bi
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Complex expressions are equal exactly when their simplified real parts and imaginary coefficients match. We’ll make the hidden coefficient on a lone imaginary unit explicit, combine like imaginary terms, retain …
Preview problemWrite the complex number represented by a point on the complex plane
Understand i as a number with i^2=-1 and represent complex numbers as a+bi.
Reading the complex plane reverses the usual plotting map: horizontal becomes the real part and vertical becomes the imaginary coefficient. We’ll read both signed coordinates, substitute them into a plus …
Preview problemAdd complex numbers in standard form by combining real parts and imaginary parts
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Complex addition is componentwise: real parts combine with real parts and imaginary coefficients combine with imaginary coefficients. We’ll regroup the two component types, add each pair independently, keep the factor …
Preview problemSubtract complex numbers in standard form by combining real parts and imaginary parts
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Complex subtraction begins by applying the minus sign to both components of the second number. We’ll distribute that sign, combine real parts and imaginary coefficients separately, restore standard form, and …
Preview problemMultiply a complex number by a real number and write the result in standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
A real scalar multiplies both coordinates of a complex number. We’ll distribute the factor to the real and imaginary terms, scale each coefficient, preserve the imaginary unit on its component, …
Preview problemMultiply a complex number by a pure imaginary number and write the result in standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Multiplying by a pure imaginary factor creates one imaginary term and one squared-imaginary term. We’ll distribute completely, replace the square of the imaginary unit with negative one, separate unlike components, …
Preview problemMultiply two complex numbers written as binomials and express the product in standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Complex-binomial multiplication follows ordinary distribution, with one crucial reduction afterward. We’ll write all four products, replace the squared imaginary unit by negative one, combine real terms and imaginary terms separately, …
Preview problemMultiply a complex number by its conjugate
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Conjugate factors share a real part and have opposite imaginary parts, causing the cross terms to cancel. We’ll recognize that structure, apply the sum-of-squares product identity created by the imaginary …
Preview problemSimplify a complex-number expression with products or powers and write the result in standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Order of operations keeps the complex product together before the separate term is combined. We’ll expand the factors, carry the added imaginary term forward, reduce the squared imaginary unit with …
Preview problemWrite the complex conjugate of a complex number in standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
A complex conjugate preserves the real part and reverses only the imaginary term’s sign. We’ll identify the two components, apply that one-sign change, keep standard form, and use the real-valued …
Preview problemAdd complex numbers geometrically on the complex plane by combining real and imaginary coordinates
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Geometric complex addition is vector addition of real and imaginary coordinates. We’ll map each number to a horizontal-vertical pair, add corresponding components, locate the resultant endpoint, translate it back to …
Preview problemDetermine whether two complex-number expressions are equivalent by simplifying to standard form
Add, subtract, and multiply complex numbers using i^2=-1 and algebraic properties.
Equivalence is easiest to test after both expressions are in standard form. We’ll group the first expression’s real parts and imaginary coefficients, simplify each component, compare those two results with …
Preview problemClassify any quadratic using the discriminant
Solve real-coefficient quadratic equations that have complex solutions.
The discriminant classifies a quadratic’s roots before the roots themselves are calculated. We’ll read the signed coefficients, substitute them carefully into b squared minus four a c, evaluate its sign, …
Preview problemSolve a quadratic of the form ax^2 + c = 0 over the complex numbers by using square roots
Solve real-coefficient quadratic equations that have complex solutions.
The missing linear term makes the square-root method direct. We’ll isolate the squared variable, take both square-root branches, separate the negative radicand into a positive square and the imaginary unit, …
Preview problemSolve a quadratic with a negative discriminant using the quadratic formula
Solve real-coefficient quadratic equations that have complex solutions.
A negative discriminant turns the quadratic formula’s radical into an imaginary term. We’ll read the signed coefficients, compute the discriminant, handle negative b carefully, rewrite the radical using i, and …
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