Figure 5 shows a container in the form of a frustum which is made by cutting off the lower part of an inverted right circular cone of base radius 48 cm and height 64 cm. The height of the container is 40 cm. The container is placed on a horizontal table. Some water is now poured into the container. Winnie finds that the area of the wet curved surface of the container is 1275 cm and the depth of water is r cm.
https://www.flickr.com/photos/130230543@N06/15919106190/
(a)Find r.
(b)Winnie claims that the volume of water in the container is less than 0.04 m3. Do you agree? Explain your answer.
If the two straight lines : y = –2x + 8 and : kx + y + 12 = 0 intersect at the
x-axis, find k.
A.–3
B.–2
C.3
D.4
(Answer is A)
The H.C.F. and the L.C.M. of three expressions are a2b2 and 120a3b5c3 respectively. If the first expression and the second expression are and respectively, then the third expression is
A.b5 c3.
B.4 a3 b5 c3.
C.40 a3 b5 c3.
D.120 b5 c3.
(Answer is C)
Suppose alpha and beta are the roots of the equation ax2 + bx + c = 0, where a, b and c are constants. If beta = 3alpha, which of the following equations must have roots -2alpha and 4alpha?
A.4ax2 - 2bx + 3c = 0
B.4ax2 + 2bx + 3c = 0
C.6ax2 - 3bx + 16c = 0
D.6ax2 + 3bx - 16c = 0
(Answer is C)
If alpha and beta are the roots of the equation x2 - 2x - 5 = 0 and alpha > beta , find the value of alpha2 + 2beta + 1.
A.–4
B.2
C.6
D.10
(Answer is D)
S4_Mock01_E
2014-12-28 5:15 am
回答 (1)
2014-12-28 7:36 am
✔ 最佳答案
r = 20Disagree, the volume is 9.0420m^3 > 0.04m^3.
2014-12-27 23:36:58 補充：
is 0.0420 m ^3
2014-12-27 23:41:40 補充：
MCQ 1,
Since y = –2x + 8 and : kx + y + 12 = 0 intersect at the x-axis.
Sub y=0 into y = –2x + 8x --> x=4, so y = –2x + 8 touch x-axis at (4,0)
From kx+y+12=0
x intercept = -12/k = 4 so k=-3
2014-12-27 23:59:53 補充：
MCQ 2 has missing data
MCQ 3:
Alpha + Beta = Alpha + 3 Alpha = 4 Alpha
Alpha x Beta = Alpha x 3 Alpha = 3(Alpha)^2
From ax^2 + bx + c = 0
Alpha + Beta = -b/a
4 Alpha = -b/a
2 Alpha = -b/2a
Alpha x Beta = c/a
3(Alpha)^2=c/a
(Alpha)^2=c/3a
2014-12-28 00:07:07 補充：
MCQ 3 con't
-8 (Alpha)^2 = -8c / 3a
The equations must have roots -2alpha and 4alpha is
x^2-(sum of root)x+(product of root) = 0
x^2 - (2 Alpha)x + [-8(alpha)^2] = 0
x^2 - (-b/2a)x + (-8c / 3a) = 0
x^2 + (b/2a)x- (8c / 3a) = 0
Both sides times 6a
6ax^2 + 3bx - 16c = 0
?? Not D
2014-12-28 00:16:57 補充：
MCQ 4
alpha and beta are the roots of the equation x^2 - 2x - 5 = 0
alpha + beta = -(-2)/1 = 2
sub x=alpha
(alpha)^2-2(alpha)-5 = 0
(alpha)^2 = 2(alpha) + 5
(alpha)^2 + 2beta + 1 = 2(alpha) + 5 + 2 beta + 1
= 2(alpha + beta) + 6
2014-12-28 00:17:53 補充：
(alpha)^2 + 2beta + 1
= 2(alpha) + 5 + 2 beta + 1
= 2(alpha + beta) + 6
= 2(2) + 6
= 10
2014-12-28 00:51:58 補充：
Let the base radius of the container be R.
R/48 = (64-40)/64
R = 18
The slant height of the cone cut out=sq rt. (18^2+24^2)=30cm
The curve surface of the cone cut out = pie (18)(30)= 540pie cm^2
So the curve surface area of the cone formed by the water and the cone cut out.
2014-12-28 00:53:49 補充：
= (540 pie + 1275pie)cm^2 = 1815pie cm^2
The slant height of the cone formed by the container and the cone cut out
=sq rt (48^2+64^2) = 80cm
The curve surface of the cone formed by the container and the cone cut out
= pie (48)(80)=3840 pie cm^2
By the ratio,
1815pie / 3840 pie =[(24+r)/64]^2
2014-12-28 01:11:20 補充：
CHECK PM
SEE THE LQ SOLUTION
NO SPACE
參考: ME!!, ME!!, ME, ME, ME, ME, ME !!, ME, ME, ME!!!
收錄日期: 2021-04-15 17:42:55
原文連結 [永久失效]:
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