Full Grade 11 test

Find the vertex of the absolute-value graph. Compare the values at the vertex and both interval endpoints. The coefficient sign determines whether the vertex is a minimum or a maximum.

Factor out x and factor the difference of squares. Find the valid part of the interval, then count its integers, excluding any root.

Perpendicular vectors have zero dot product. Expand (u + tv) · v and remember that v · v is the squared length.

From b√x + a = 0, obtain √x = −a/b for b ≠ 0. Require x ≥ 0 and −a/b ≥ 0. Then square: x = (−a/b)². Check the original equation.

Change the base to a or to e. Bring the exponents outside the logarithms, keeping track of the coefficients in numerator and denominator.

The intersection A ∩ B contains elements in both sets. Count shared elements once in the union: |A ∪ B| = |A| + |B| − |A ∩ B|. Thus intersection size = |A| + |B| − |A ∪ B|, and |A \ B| = |A| − |A ∩ B|.

Sine and cosine have period 360°; tangent has period 180°. sin(−α) = −sin α. Sine is positive in quadrants I and II and negative in III and IV; cosine is also negative in III. Special values: sin 30° = 1/2, sin 45° = √2/2, sin 60° = √3/2.

Graphical solutions of f(x) = a are the x coordinates where the graph meets y = a. Algebra often gives exact values while a graph generally gives approximations. Check every intersection and the specified interval.

Solve the vector equation separately for each coordinate. Subtract the known multiple of v and divide by the coefficient of u.

The square root increases with its radicand. Evaluate both endpoints. A negative coefficient reverses monotonicity; list the range endpoints in increasing order.

A percentage-increase multiplier must include both the entire starting value and its increase.

1. The starting value is 100%, and the increase adds another p%. The new value is therefore (100 + p)% of the starting value.
2. Convert this percentage into a decimal by dividing by 100.
3. The multiplier is q = (100 + p) / 100 = 1 + p / 100.

Multiplying by p / 100 alone gives only the increase, not the whole new value. Check that an increase multiplier is greater than 1.

Calculate differences between several consecutive terms and, when the preceding term is nonzero, their ratios. A constant difference indicates an arithmetic progression; a constant ratio indicates a geometric progression. Check every pair.

Set t = x², requiring t ≥ 0. Solve t² − st + p = 0. Discard negative t; each positive t gives x = ±√t, and t = 0 gives only x = 0.

A square root is defined over the real numbers only when its radicand is non-negative.

1. Write the condition that the entire expression under the square root must be at least zero.
2. Solve the resulting inequality by moving the constant term to the other side. Pay attention to parentheses: subtracting a negative number means adding it.
3. Check the boundary value and determine on which side the radicand is non-negative. Zero under the square root is allowed, so the boundary value is included in the domain.

Expand (u + v)³ and group the mixed terms by factoring out uv. Express the sum of cubes using the given sum and product.

An arithmetic progression has difference d = a₂ − a₁. Its nth term is aₙ = a₁ + (n − 1)d. The first n terms sum to Sₙ = n(a₁ + aₙ)/2, or n(2a₁ + (n − 1)d)/2.

Every logarithm argument must be > 0, with base a > 0 and a ≠ 1. logₐ u = r means u = aʳ. Equal logarithms of the same base have equal positive arguments. Check all arguments and the original equation after finding x.

An even power removes a negative factor sign; an odd power preserves it. Check factor signs in each interval and count negative products.

An arithmetic progression has difference d = a₂ − a₁. Its nth term is aₙ = a₁ + (n − 1)d. The first n terms sum to Sₙ = n(a₁ + aₙ)/2, or n(2a₁ + (n − 1)d)/2.

Every logarithm argument must be > 0, with base a > 0 and a ≠ 1. logₐ u = r means u = aʳ. Equal logarithms of the same base have equal positive arguments. Check all arguments and the original equation after finding x.

Convert the first inequality into a double inequality using distance. Intersect the two conditions. Check the inclusion of each endpoint separately.

Equal absolute values mean the expressions inside are equal or opposite. Form two linear equations, check both solutions, and perform the requested operation.

Each year retains a fraction 1 − p/100 of that year starting amount. After n years the value is S × (1 − p/100)ⁿ. Apply each decrease to the already reduced amount.

A logarithmic base between zero and one makes the function decreasing; a base above one makes it increasing. A negative coefficient reverses the direction. Check positive arguments and compare the inputs.

Determine monotonicity from the base. Evaluate the interval endpoints, accounting for both shifts. Include the endpoint values because the interval is closed.

The intersection A ∩ B contains elements in both sets. Count shared elements once in the union: |A ∪ B| = |A| + |B| − |A ∩ B|. Thus intersection size = |A| + |B| − |A ∪ B|, and |A \ B| = |A| − |A ∩ B|.