Full Grade 8 test

Vector AB equals endpoint coordinates minus starting coordinates: (xB − xA, yB − yA). Order matters: BA is the opposite vector.

A difference of squares factors as a² − b² = (a − b)(a + b). Apply this to the given bases. Expanding the product cancels the cross terms.

Treat the longest side as c and check a² + b² = c². If this holds, the converse Pythagorean theorem shows the triangle is right-angled.

Interquartile range IQR = Q3 − Q1. It describes the spread of the middle approximately 50% of the data. A greater IQR means greater spread in this middle portion.

The pyramid face slant height, vertical height h and half the base side a/2 form a right triangle. Slant height = √(h² + (a/2)²). Use half the side, not the full side.

Subtract vectors coordinate by coordinate: (u₁, u₂) − (v₁, v₂) = (u₁ − v₁, u₂ − v₂).

1. Subtract the first coordinate of the second vector from the first coordinate of the first vector.
2. Do the same with the second coordinates.
3. Write the two results as a new ordered pair.

Take particular care with signs: subtracting a negative coordinate becomes addition.

First find the base diagonal e = √(a² + b²). Then the space diagonal d = √(e² + h²), giving d = √(a² + b² + h²). Apply Pythagoras twice.

Substitute the given solution for x. Square it first, multiply by the coefficient, then isolate c in the resulting equality.

The intersection A ∩ B contains only elements belonging to both sets. Compare the lists and keep the common elements; elements appearing in only one set do not qualify.

The route with a turn is a + b; the direct route is √(a² + b²). When both lengths are positive, the direct route is shorter. If a difference is needed, subtract the direct length from a + b.

Sort the data. Q2 is the overall median; Q1 is the median of the lower half and Q3 the median of the upper half. In these tasks, an even-length list is split into two equal halves; the median of an even-sized group is the average of its two middle values.

Subtract the constant from all three parts, then divide all parts by the positive coefficient. Exclude a strict boundary and include a non-strict one.

Relative frequency percentage = interval frequency ÷ total frequency × 100. For example, 6 out of 20 is 30%.

A polynomial with one, two or three terms is a monomial, binomial or trinomial respectively. The expression ax² + bx + c has three terms, so it is a trinomial.

In a right triangle, the leg opposite 30° is half the hypotenuse. Divide the given hypotenuse by 2.

A difference of squares factors as a² − b² = (a − b)(a + b). Apply this to the given bases. Expanding the product cancels the cross terms.

The opposite vector has the same magnitude and reversed direction. Both coordinates change sign: (a, b) → (−a, −b).

Square of a difference: (x − a)² = x² − 2ax + a². The middle term is negative and the final term positive because (−a) × (−a) = a².

The magnitude of (a, b) is √(a² + b²), by Pythagoras. Magnitude is always nonnegative.

With equal interval widths, the tallest histogram bar gives the greatest frequency. Select the interval beneath that bar.

Perpendicular lengths form the legs of a right triangle. By Pythagoras, hypotenuse c = √(a² + b²). Square the lengths, add and take the square root; do not simply add the lengths.

Adding x + y = s and 2x − y = t eliminates y: 3x = s + t. Find x, then use y = s − x.

First calculate IQR = Q3 − Q1. The upper outlier fence is Q3 + 1.5 × IQR. A value is an upper outlier if it is strictly greater than this fence.

Multiply every coordinate by k: k(a, b) = (ka, kb). A negative k also reverses the direction.