Full Grade 10 test

A fraction is zero when its numerator is zero and denominator nonzero. The candidate is x = r, but it must satisfy x ≠ d. If r = d, the candidate is invalid and there is no solution.

Rectangle side lengths sum to half the perimeter and multiply to the area. Form t² − (P/2)t + S = 0. Positive roots give the sides; check both perimeter and area.

Opposite angles of a cyclic quadrilateral total 180°. Missing angle = 180° − given opposite angle.

For c > 0, x² < c means −√c < x < √c. For x² > c, take x < −√c or x > √c. Strict inequalities exclude the boundary values.

A point satisfies a system of equations only when its coordinates satisfy both equations.

1. Substitute the point’s x-coordinate into the first equation and calculate x².
2. Compare the result with the point y-coordinate.
3. Also check that the same y-coordinate satisfies the second equation.

If either equality is false, the point does not satisfy the system.

Cosine rule: c² = a² + b² − 2ab cos C. With three sides, cos C = (a² + b² − c²)/(2ab), then use arccos. For C = 90°, cos C = 0, giving Pythagoras.

Dividing a total V in ratio a:b gives a + b equal units. First share = V × a/(a + b), second share = V × b/(a + b). Check that they sum to V.

Equate the y expressions: x² = c. For c > 0 there are two pairs (−√c, c) and (√c, c); for c = 0 one pair (0, 0); for c < 0 no real solutions.

An angle bisector divides an angle into two equal parts. It divides the opposite side proportionally to the adjacent sides: x/X = a/b. With that correspondence, X = xb/a.

For successive choices joined by “and”, multiply: shirt and trousers give a × b outfits. For disjoint alternatives joined by “or”, add: bus or train gives a + b choices.

Use vertex form y = a(x − m)² + n.

1. Insert the vertex coordinates m and n.
2. Substitute the coordinates of A for x and y.
3. Solve the resulting equation for a.
4. Check that substituting the point’s x-coordinate into the selected formula gives its y-coordinate.

If x + y = s and xy = p, both numbers are roots of t² − st + p = 0. Two distinct roots give two ordered pairs by swapping them. If x < y is required, put the smaller first; equal roots give one pair.

Order the roots a < b. The product (x − a)(x − b) is positive outside the roots, negative between them and zero at the roots. Strict > and < exclude roots; ≤ and ≥ include them. Select the required intervals.

Cosine rule: c² = a² + b² − 2ab cos C. With three sides, cos C = (a² + b² − c²)/(2ab), then use arccos. For C = 90°, cos C = 0, giving Pythagoras.

Order the roots a < b. The product (x − a)(x − b) is positive outside the roots, negative between them and zero at the roots. Strict > and < exclude roots; ≤ and ≥ include them. Select the required intervals.

Adding the same constant to all values leaves spread and variance unchanged. Multiplying all values by k multiplies standard deviation by |k| and variance by k². A larger standard deviation means greater spread.

In the strict double inequality a < x < b, x must be greater than a and less than b. The boundary values themselves are excluded.

1. Find the smallest integer strictly greater than the lower bound.
2. Find the largest integer strictly less than the upper bound.
3. Count all integers from the first to the last, inclusive: last − first + 1. If the first exceeds the last, there are no solutions.

When both bounds a and b are integers and a < b, the permitted integers run from a + 1 to b − 1, so their number is b − a − 1. Remember that 0 is also an integer and must be counted if it lies inside the interval.

Triangle area is S = r × p, where r is the inradius and p the semiperimeter. Thus r = S/p. Do not confuse semiperimeter with full perimeter.

With order important and no repetition, arranging n elements gives n! possibilities and assigning r distinct roles gives n!/(n − r)!. An unordered pair gives n(n − 1)/2. A code with r positions, n symbols per position and repetition allowed gives nʳ.

Similar figures have equal corresponding angles and proportional sides. Lengths and perimeters scale by k; areas by k². Find k as target length divided by original length. Multiply area by k² to enlarge and divide to shrink. Sides 4 and 12 give k = 3 and an area factor of 9.

For c > 0, x² < c means −√c < x < √c. For x² > c, take x < −√c or x > √c. Strict inequalities exclude the boundary values.

The golden ratio is φ ≈ 1.618: longer part / shorter part = φ. Multiply the shorter side by 1.618 to find the longer; divide the longer by 1.618 to find the shorter. Round the final result to the precision of the choices.

Similar figures have equal corresponding angles and proportional sides. Lengths and perimeters scale by k; areas by k². Find k as target length divided by original length. Multiply area by k² to enlarge and divide to shrink. Sides 4 and 12 give k = 3 and an area factor of 9.

With order important and no repetition, arranging n elements gives n! possibilities and assigning r distinct roles gives n!/(n − r)!. An unordered pair gives n(n − 1)/2. A code with r positions, n symbols per position and repetition allowed gives nʳ.

The pair values are x and y. Calculate their sum and product separately. The pair works only when both x + y = s and xy = p hold.

Complement probability = 1 − P(A), or 100% − p% in percentages. In the finite models used here, 0 denotes an impossible event and 1 a certain event.

Select the table columns satisfying the event and add their frequencies. Divide by the sum of all frequencies. For the theoretical probability, count favourable faces out of six equally likely faces. Subtract the theoretical probability from the empirical one and check the sign.

The vertex of a downward-opening parabola represents the maximum height.

1. From vertex form, read that the maximum height is n and occurs after m seconds.
2. Set h(t) equal to zero to find when the ball lands.
3. Solve the equation and choose the later non-negative time.

Similar figures have equal corresponding angles and proportional sides. Lengths and perimeters scale by k; areas by k². Find k as target length divided by original length. Multiply area by k² to enlarge and divide to shrink. Sides 4 and 12 give k = 3 and an area factor of 9.

Equate the y expressions: x² = c. For c > 0 there are two pairs (−√c, c) and (√c, c); for c = 0 one pair (0, 0); for c < 0 no real solutions.

Adding the same constant to all values leaves spread and variance unchanged. Multiplying all values by k multiplies standard deviation by |k| and variance by k². A larger standard deviation means greater spread.

Fibonacci rule: each term from the third onward is the sum of the two preceding terms. In 1, 1, 2, 3, 5, ... count positions from the first term. For a missing middle term, subtract the earlier term from the following term.

Complement probability = 1 − P(A), or 100% − p% in percentages. In the finite models used here, 0 denotes an impossible event and 1 a certain event.

From x² = ax obtain x(x − a) = 0, so x = 0 or x = a. Find y = x² for each x and give the solution pairs.