問題一覧
1
25
2
x = 0 or 2
3
60/143 minutes gain
4
Using the rule that powers multiply by adding their exponents, a multiplicative magic square is easily obtained from the given square by substituting 2^n (or k^n where k > 1) for n in each block
5
Equal to 4
6
4:26.853
7
T1二0^∞ 【x^n=1/(1-x)】
8
Pat = 8 errors, Mike = 2 errors
9
since each v/w^2, w^3/x^4, x^5/y^6 and y^7/z^4 is a constant, therefore their product is also a constant
10
12
11
33 pearls
12
1.38 ft or 3.62 ft
13
Jones = 0.5 hours, Smith = 3.5 hours
14
1414
15
764,488
16
pencil = 26, eraser = 19, notebook = 55
17
The two expressions are identically equal, respectively, to the smaller and the larger of the two numbers x and y.
18
25%
19
A = 9/4, B = 27/8
20
50 rupees
21
80 steps
22
The number of solutions of a set of polynomial equations is the product of their degrees. Since hyperbolas are of 2nd degree, the solutions is to a pair of hyperbolic equations cannot exceed 4. Thus, two hyperbolas can intersect in no more than 4 points.
23
50 miles/hour, 180 miles
24
t2+1/2[(T2-T1)-(t2-t1)]
25
29
26
10
27
50 flags
28
97
29
5562
30
V1 = ¾ V2
31
Same
32
2:54:35 and 9:05:25
33
Place five at the vertices of a regular pentagon, the sixth at the center of the pentagon, and the seventh above the center at a distance equal to the radius of the pentagon.
34
17.9 miles
35
It suffices to prove it for two sacred numbers, since the theorem then follows by induction. Let M = A^2 + B^2 and N = C^2 + D^2. Then MN = (A^2 + B^2)(C^2 + D^2) = (AC + BD)^2 + (AD - BC)^2. Amen.
36
From P draw a line, L, to the opposite vertex, say A. Now construct a line parallel to L from the midpoint of BC, intersecting the side of the triangle at Q. The line PQ divides the triangle into two equal areas.
37
a^x+b^x=a^2 (a^(x-2))+b^2 (b^(x-2)) < a^2 c^(x- 2) +b^2 c^(x-2)=(a^2+b^2) c^(x-2) =c^2 c^(x-2)=c^x
38
24 inches
39
100 yards
40
One cut in the third link will allow two links to be swapped for a kiss and a link on the second transaction, and 3 links for a kiss and 2 links on the third and so on.
41
1437.45 ft
42
58
43
no, outside the triangle
44
2% down from the top of cup
45
sqrt(18)
46
Equal
47
2 days
48
20.69 inches
49
1
50
a=25,b=16,c=39
51
2.5 in^2
52
5pi
53
600 yards
54
North pole, (sqrt(2))×10^7 meters
55
14
56
The angle a = pi/10 satisfies the relation tan 3a = cot 2a
57
Area = 939120, Length = 2018
58
one acre
59
Draw two lines through the center of the square, perpendicular to each other, such that each is parallel to a side of the unit square. These two lines partition the unit square into four ½ unit squares. At least 2 of the 5 points must be in (or on the perimeter of) one of these smaller squares. This pair of points cannot be farther apart than the length of the small square's diagonal sqrt(2/2) units
60
13 ft. and 5 inches
61
R= 438.14
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Definition of Terms 1
90問 • 1年前Definition of Terms 2
Definition of Terms 2
Justin · 90問 · 1年前Definition of Terms 2
Definition of Terms 2
90問 • 1年前Definition of Terms 3
Definition of Terms 3
Justin · 90問 · 1年前Definition of Terms 3
Definition of Terms 3
90問 • 1年前問題一覧
1
25
2
x = 0 or 2
3
60/143 minutes gain
4
Using the rule that powers multiply by adding their exponents, a multiplicative magic square is easily obtained from the given square by substituting 2^n (or k^n where k > 1) for n in each block
5
Equal to 4
6
4:26.853
7
T1二0^∞ 【x^n=1/(1-x)】
8
Pat = 8 errors, Mike = 2 errors
9
since each v/w^2, w^3/x^4, x^5/y^6 and y^7/z^4 is a constant, therefore their product is also a constant
10
12
11
33 pearls
12
1.38 ft or 3.62 ft
13
Jones = 0.5 hours, Smith = 3.5 hours
14
1414
15
764,488
16
pencil = 26, eraser = 19, notebook = 55
17
The two expressions are identically equal, respectively, to the smaller and the larger of the two numbers x and y.
18
25%
19
A = 9/4, B = 27/8
20
50 rupees
21
80 steps
22
The number of solutions of a set of polynomial equations is the product of their degrees. Since hyperbolas are of 2nd degree, the solutions is to a pair of hyperbolic equations cannot exceed 4. Thus, two hyperbolas can intersect in no more than 4 points.
23
50 miles/hour, 180 miles
24
t2+1/2[(T2-T1)-(t2-t1)]
25
29
26
10
27
50 flags
28
97
29
5562
30
V1 = ¾ V2
31
Same
32
2:54:35 and 9:05:25
33
Place five at the vertices of a regular pentagon, the sixth at the center of the pentagon, and the seventh above the center at a distance equal to the radius of the pentagon.
34
17.9 miles
35
It suffices to prove it for two sacred numbers, since the theorem then follows by induction. Let M = A^2 + B^2 and N = C^2 + D^2. Then MN = (A^2 + B^2)(C^2 + D^2) = (AC + BD)^2 + (AD - BC)^2. Amen.
36
From P draw a line, L, to the opposite vertex, say A. Now construct a line parallel to L from the midpoint of BC, intersecting the side of the triangle at Q. The line PQ divides the triangle into two equal areas.
37
a^x+b^x=a^2 (a^(x-2))+b^2 (b^(x-2)) < a^2 c^(x- 2) +b^2 c^(x-2)=(a^2+b^2) c^(x-2) =c^2 c^(x-2)=c^x
38
24 inches
39
100 yards
40
One cut in the third link will allow two links to be swapped for a kiss and a link on the second transaction, and 3 links for a kiss and 2 links on the third and so on.
41
1437.45 ft
42
58
43
no, outside the triangle
44
2% down from the top of cup
45
sqrt(18)
46
Equal
47
2 days
48
20.69 inches
49
1
50
a=25,b=16,c=39
51
2.5 in^2
52
5pi
53
600 yards
54
North pole, (sqrt(2))×10^7 meters
55
14
56
The angle a = pi/10 satisfies the relation tan 3a = cot 2a
57
Area = 939120, Length = 2018
58
one acre
59
Draw two lines through the center of the square, perpendicular to each other, such that each is parallel to a side of the unit square. These two lines partition the unit square into four ½ unit squares. At least 2 of the 5 points must be in (or on the perimeter of) one of these smaller squares. This pair of points cannot be farther apart than the length of the small square's diagonal sqrt(2/2) units
60
13 ft. and 5 inches
61
R= 438.14