Answer:
27) 18 < P ≤ 18 + 6√2 ⇒ answer D
28) The sum of the degree measures these angles is 1080° ⇒ answer B
29) 3E minutes before A ⇒ answer B
30) The difference between the greatest possible values is 0 ⇒ answer E
31) r divided by s = 1/3 ⇒ answer A
Step-by-step explanation:
* Lets explain each problem
27)
∵ BE is a quarter circle
∵ The radius of the circle is 6
∵ Point c is on the arc BE
∴ The distance from D to C = 6 ⇒ not depends on the position of c
because DC is a radius in the quarter circle BE
- In Δ BDE
∵ m∠ D = 90°
∵ DB = DE = 6 ⇒ radii of the quarter circle
- By using Pythagoras Theorem
∴ BE = √ (6² + 6²) = √(36 + 36) = √72 = 6√2
- The perimeter of the quadrilateral ABCD is the sum of the sides
∵ AB = 6 , AD = 6 , CD = 6
- Point C can move from B to E
∴ The length of side BC can b greater than 0(it can not be 0
because the quadrilateral has 4 sides
∴ The length of BC can not exceed the length of BE because the last
position of point C to be on the arc BE is point E
∴ The length of BC ⇒ 0 < BC ≤ 6√2
equal 6√2
∵ P is the perimeter of the quadrilateral ABCD
∴ P = 6 + 6 + 6 + (0 < BC ≤ 6√)
∴ P = 18 + (0 < BC ≤ 6√)
- Add 18 to 0 and 18 to 6√2
∴ 18 < P ≤ 18 + 6√2
28)
- In the figure we have a quadrilateral
- All the arrows represent the exterior angles of the figures
- Use the fact that:
The sum of all angles around a points is 360°
∵ There are 4 vertices (points) on the quadrilateral
∴ The sum of the all angles around the 4 vertices = 4 × 360 = 1440°
- Use the fact that:
The sum of the interior angles of any quadrilateral is 360°
∵ The sum of the angles represented by the arrows is the difference
between the sum of all angles around the 4 vertices and the sum
of the interior angles of the quadrilateral
∴ The sum of these angles = 1440° - 360° = 1080°
* The sum of the degree measures these angles is 1080°
29)
- In any watch the short arrow-hand represents the hours and the long
arrow-hand represents the minutes
- The numbers of the hours in the watch from 1 to 12
- The number of minutes between each two hours is 5 minutes, then
at 1 o'clock the minutes number is 5 , at 6 o'clock the number of
minutes is 30 , at 9 o'clock the number of minutes is 45 , so we can
find the number of minutes at any number of hour by multiply the
number of hour by 5
∵ The number of hours have been replaced by letters
∵ The time on the watch is 45 minutes after 12 o'clock <em>OR</em>
15 minutes before 1 o'clock
∵ The short arrow-hand pointed between L and A
∵ L is the replacing of 12 o'clock and A is the replacing of 1 o'clock
∵ The long arrow-hand pointed at I
∵ I is the replacing of 9 o'clock
∵ The hour number 9 means 5 × 9 = 45 minutes
∴ The hour hand I has 5I minutes
∴ The time in letter is 5I minutes after L
- This answer is not in the choices
- But the answer of 3E minutes before A means:
∵ E is the replacing of 5 o'clock
∴ 3E = 3 × 5 = 15 minutes
∵ A is the replacing of 1 o'clock
∴ 3E minutes before A means 15 minutes before 1 o'clok
* The answer is ⇒ 3E minutes before A
30)
∵ r² = 9
∴ r = ± √9 = ± 3
∴ r has two values -3 and 3
∵ s² = 25
∴ s = ± √25 = ± 5
∴ s has two values -5 and 5
- To find the greatest value of s - r put s greatest and r smallest
∵ The greatest value of s is 5
∵ The smallest value of r is -3
∴ The greatest value of s - r = 5 - (-3) = 5 + 3 = 8
- To find the greatest value of r - s put r greatest and s smallest
∵ The greatest value of r is 3
∵ The smallest value of s is -5
∴ The greatest value of r - s = 3 - (-5) = 3 + 5 = 8
∴ The difference between the greatest possible values of s - r
and r - s = 8 - 8 = 0
* The difference between the greatest possible values is 0
31)
- There are 27 cubes each of side length r
- The 27 cubes are arranged to form on single large cube of side
length s
∵ The volume of any cube is V = L³ , where L is the length of its side
∵ The large cube formed from the 27 small cubes
∴ The volume of the large cube = the volume of the 27 small cubes
∵ The side of the small cube is r
∴ The volume of the small cube is r³
∵ The side of the large cube is s
∴ The volume of the large cube is s³
∴ s³ = 27 r³
- Divide both sides by s³ and 27
∴ s³/(27 s³) = (27 r³)/(27 s³)
∴ 1/27 = r³/s³
- Take ∛ for both sides
∴ ∛(r³/s³) = ∛(1/27)
- The cube root canceled by the power 3 and the cube root of
1/27 is 1/3
∴ r/s = 1/3
* r divided by s = 1/3
