Full Grade 10 test

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.

A fair die has six equally likely outcomes. Count outcomes from 1 through n and divide by 6. For 1 ≤ n ≤ 6, the probability is n/6.

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.

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.

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.

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.

Adding only water leaves the pure substance mass unchanged. Initially it is m × p/100. At 10% concentration the total mass must equal this amount / 0.10. Water added = new total mass − original mass.

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.

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.

Pure substance mass = total mass × its percentage / 100. If p is the water percentage, the dry matter percentage is 100 − p, so its mass = total mass × (100 − p)/100.

A unit-circle point has coordinates (cos α, sin α). In quadrant II, x is negative and y positive, so cosine is negative and sine positive. Use special-angle sine and cosine values for coordinates.

All denominators in a rational expression must be nonzero. Denominator x − a excludes x = a. After solving, discard forbidden values and substitute the remaining candidates into the original equation.

Count unordered pairs from the first group: swapping the pair does not create a new choice. Count choices for the second-group member separately. Both selections are required, so multiply their counts.

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.

Adding only water leaves the pure substance mass unchanged. Initially it is m × p/100. At 10% concentration the total mass must equal this amount / 0.10. Water added = new total mass − original mass.

Square the given x to find y. Give the solution as (x, x²), with x first.

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.

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.

Midpoint coordinates are the averages of endpoint coordinates: ((x1 + x2)/2, (y1 + y2)/2). Calculate each coordinate separately.

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.

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.

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.

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 unit-circle point has coordinates (cos α, sin α). In quadrant II, x is negative and y positive, so cosine is negative and sine positive. Use special-angle sine and cosine values for coordinates.

Triangle area S = ab sin C / 2, with C between sides a and b. For fixed a and b, the greater sine gives the greater area.