1.arrange the following in descanding order:
0.087X 10^-12, 0.11X 10^-15, 42.8X10^-12
2.the ratio of the number of marbles by Susan to the marblels owned by Theresa is 5:2. susan has n marbles. if susan gives 18 of her own marbles to Theresa, both of them will have the same number of marbles. find n.
3.the ratio of the volume of two simliar solid cones is 8:27.
a. find the ratio of the height of the smaller cone to the height of the larger one.
b. if the cost of painting a cone varies as its total surface are and the cost of painting the smaller cone is $32, find the cost of painting the larger cone.
4. which of the following are the possible lengths of a right-angled triangle?
A.10,24,26 B.10,25,26 C.11,25,26 D.12,25,27
f.2 math題
2011-08-21 2:56 am
回答 (2)
2011-08-22 9:41 am
✔ 最佳答案
1.0.087X 10^-12 = 8.7 X 10^-14
0.11X 10^-15 =1.1 X 10^-16
42.8X10^-12 = 4.28 X 10^-11
42.8X10^-12, 0.087X 10^-12, 0.11X 10^-15 (in descending order)
2.
Let n = number of marbles Susan owns
Theresa/Susan= 2/5, Theresa = 2/5 Susan
Theresa own = (2/5) n
n – 18 = (2/5) n + 18
n - (2/5) n =18 + 18
3/5 n = 36
n = 60
3.
(a)
Pi(r1)^2(h1) / Pi(r2)^2(h2) = 8/27
(r1)^2(h1) / (r2)^2(h2) = 8/27
h1/h2= 8(r2)^2/27(r1)^2 -------- (1)
r2/r1 = h2/h1 (similar solid cones)
[r2/r1]^2 =[ h2/h1]^2 ----- (2)
substitute (2) into equation (1)
h1/h2= 8(h2)^2/27(h1)^2
h1 (8) (h2)^2
---- = ------------------
h2 (27)(h1)^2
(h1)^3 8
---------- = -------
(h2)^3 27
Take the cubic root
h1/h2 = 2/3
(b)
Suppose big cone: r2 = 3 m, h2 = 4 m, s2(slant height) = 5m
small cone: r1 = 2 m, h1 = 8/3 m, s1(slant height) = 10/3 m
Surface area of big cone A2 = pi r1 s1 + pi r1^2 = pi (3)(5) + pi (3)^2 = 26 pi m^2
Surface area of small cone A1= pi r2 s2 + pi r2^2 = pi (2)(10/3) + pi (2)^2 = 32/3 pi m^2
A2/A1 = C2/C1 (C2 = cost of painting big cone, C1 = cost of painting small cone)
C2 = C1(A2/A1) = $32(A2/A1) = $32(26 pi)/[(32/3)pi] = $32(26 )(3/32) = $78
The cost of painting the larger cone = $78
4.Which of the following are the possible lengths of a right-angled triangle?
A.10,24,26 B.10,25,26 C.11,25,26 D.12,25,27
Only A. 10, 24, 26 is a right-angled triangle
(26^2 = 20^2 + 24^2, 679 = 579 + 100, 679 = 679 ~ Pythagorean Theorem)
2011-08-29 2:25 am
1.
0.087X 10^-12 = 8.7 X 10^-14
0.11X 10^-15 =1.1 X 10^-16
42.8X10^-12 = 4.28 X 10^-11
42.8X10^-12, 0.087X 10^-12, 0.11X 10^-15 (in descending order)
2.
Let n = number of marbles Susan owns
Theresa/Susan= 2/5, Theresa = 2/5 Susan
Theresa own = (2/5) n
n – 18 = (2/5) n + 18
n - (2/5) n =18 + 18
3/5 n = 36
n = 60
3.
(a)
Pi(r1)^2(h1) / Pi(r2)^2(h2) = 8/27
(r1)^2(h1) / (r2)^2(h2) = 8/27
h1/h2= 8(r2)^2/27(r1)^2 -------- (1)
r2/r1 = h2/h1 (similar solid cones)
[r2/r1]^2 =[ h2/h1]^2 ----- (2)
substitute (2) into equation (1)
h1/h2= 8(h2)^2/27(h1)^2
h1 (8) (h2)^2
---- = ------------------
h2 (27)(h1)^2
(h1)^3 8
---------- = -------
(h2)^3 27
Take the cubic root
h1/h2 = 2/3
(b)
Suppose big cone: r2 = 3 m, h2 = 4 m, s2(slant height) = 5m
small cone: r1 = 2 m, h1 = 8/3 m, s1(slant height) = 10/3 m
Surface area of big cone A2 = pi r1 s1 + pi r1^2 = pi (3)(5) + pi (3)^2 = 26 pi m^2
Surface area of small cone A1= pi r2 s2 + pi r2^2 = pi (2)(10/3) + pi (2)^2 = 32/3 pi m^2
A2/A1 = C2/C1 (C2 = cost of painting big cone, C1 = cost of painting small cone)
C2 = C1(A2/A1) = $32(A2/A1) = $32(26 pi)/[(32/3)pi] = $32(26 )(3/32) = $78
The cost of painting the larger cone = $78
4.Which of the following are the possible lengths of a right-angled triangle?
A.10,24,26 B.10,25,26 C.11,25,26 D.12,25,27
Only A. 10, 24, 26 is a right-angled triangle
(26^2 = 20^2 + 24^2, 679 = 579 + 100, 679 = 679 ~ Pythagorean Theorem)
0.087X 10^-12 = 8.7 X 10^-14
0.11X 10^-15 =1.1 X 10^-16
42.8X10^-12 = 4.28 X 10^-11
42.8X10^-12, 0.087X 10^-12, 0.11X 10^-15 (in descending order)
2.
Let n = number of marbles Susan owns
Theresa/Susan= 2/5, Theresa = 2/5 Susan
Theresa own = (2/5) n
n – 18 = (2/5) n + 18
n - (2/5) n =18 + 18
3/5 n = 36
n = 60
3.
(a)
Pi(r1)^2(h1) / Pi(r2)^2(h2) = 8/27
(r1)^2(h1) / (r2)^2(h2) = 8/27
h1/h2= 8(r2)^2/27(r1)^2 -------- (1)
r2/r1 = h2/h1 (similar solid cones)
[r2/r1]^2 =[ h2/h1]^2 ----- (2)
substitute (2) into equation (1)
h1/h2= 8(h2)^2/27(h1)^2
h1 (8) (h2)^2
---- = ------------------
h2 (27)(h1)^2
(h1)^3 8
---------- = -------
(h2)^3 27
Take the cubic root
h1/h2 = 2/3
(b)
Suppose big cone: r2 = 3 m, h2 = 4 m, s2(slant height) = 5m
small cone: r1 = 2 m, h1 = 8/3 m, s1(slant height) = 10/3 m
Surface area of big cone A2 = pi r1 s1 + pi r1^2 = pi (3)(5) + pi (3)^2 = 26 pi m^2
Surface area of small cone A1= pi r2 s2 + pi r2^2 = pi (2)(10/3) + pi (2)^2 = 32/3 pi m^2
A2/A1 = C2/C1 (C2 = cost of painting big cone, C1 = cost of painting small cone)
C2 = C1(A2/A1) = $32(A2/A1) = $32(26 pi)/[(32/3)pi] = $32(26 )(3/32) = $78
The cost of painting the larger cone = $78
4.Which of the following are the possible lengths of a right-angled triangle?
A.10,24,26 B.10,25,26 C.11,25,26 D.12,25,27
Only A. 10, 24, 26 is a right-angled triangle
(26^2 = 20^2 + 24^2, 679 = 579 + 100, 679 = 679 ~ Pythagorean Theorem)
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