1. The equation of straight line L is x-2y+6=0. The coordinates of point P are (-2, -3). Point Q lies on L such that PQ is perpendicular to L. Find
a) the x-intercept and the y-intercept of L
b) the equation of straight linr PQ
c) the coordinates of point Q
d) the lenght of line segment PQ (Leave the radical sign √ in the answer)
2.PQRS is a rectangle. The coordinates of point P are (2, 1 ). If the equation of staright line PS is y=1/2 x, find the equation of PQ,QR,RS and the coordinates of point Q,R,S
3.If the roots of x^2+ax+12=0 are integers, what are the possible values of a?
4. The figure shows a rectengular field ABCD where path AB is shorter than path BC. Tom and Ken are jodding from point A to point C along two different routes but with the same speed 2m/s. Tom takes 35 seconds to run from A to B and then from B to C. On the other hand, Ken runs along a straight path from A to C. If Ken reaches C 10 seconds earlier than Tom, find the lenght of AB.
5.If one root of the equation X^2-(2k-5)x+k^2-5k+56/9=0, where k is a constant, is twice the other root,
a)prove that one root is (2k-5)/3
b)find the two possible values of k
6. Andy is going to hold a concert at the Hong Kong Coliseum. Suppose the auditorium can accommodate at most 12000 people. He estimates that if the price for each ticket is $160 , then all the tickets will be sold. For every increase of $1 in the ticket price, the number of tickets sold will decrease by 50
a) Let $p be the price os each ticket and q be the number of tivkets sold. Express q in trems of p.
b) What should the price of each ticket be in order to get the maximum income.
7a) Find the remainder of (x^2008+3)/(x+1)
b) If today is Saturday, what day of the week is the day after(6^2008+3)days?
F.4 MATHS
2014-02-23 4:36 am
回答 (2)
2014-02-26 8:29 am
✔ 最佳答案
1)a)
x-intercept : Sub y = 0 into L; x= -6
y-intercept : Sub x = 0 into L; y = 3
b)
By equation of L:
x - 2y + 6 = 0
y = (1/2) x + 3
So, slope of L = (1/2)
Note that L perpendicular to PQ at Q
So, (Slope of PQ) x (Slope of L) = -1
So, Slope of PQ = -2
Equation of PQ:
(y+3) = (-2) (x+2)
Ans: 2x + y + 7 =0
c) Interception of L and PQ
Sub equation L into 2x + y + 7 = 0
Get answers: Q (-4, 1)
d) SQRT( (-2 +4)^2 + (-3 - 1)^2 )
SQRT (20)
Ans: 2 x SQRT(5)
2014-02-26 00:38:41 補充：
2. 如果 P (2, 1), 甘 PS y = 1/2 x 並唔合理
3. Delta of x^2+ax+12=0
(a)^2 - 4 (1) (12) >= 0
a >= 6.92
So, possible values of a >= 7, for a is integer
4.
Length of AB = 2 x 35 = 70 m
Let length of BC be x
Time taken for Ken to reach C = [35 + (x/2) - 10] seconds
cont...
2014-02-26 00:43:26 補充：
Length of AC = SQRT( AB^2 + BC^2)
[35 + (x/2) - 10] x 2 = SQRT ( 70^2 + x^2)
x = 24
Length of BC is 24 m, AB is 70 m
5. Let one root be Q, such that another root is 2Q
Sum of roots = 3Q = 2k-5 -----------(1)
Product of roots = 2Q^2 = k^2-5k+56/9 -----------(2)
(2) / (1):
Q = (2k-5)/3
2014-02-26 00:51:17 補充：
b) Sub k to find answer
6) 題目不合理
7)
a)
Let f(x) = (x^2008+3)
Sub f(1) => 4
By Reminder Theorem, remainder is 4
b) two days after Saturday, So it's Monday
2014-02-23 5:07 am
又傻豬左啦～
20 分 問 七題，有人答你的機會很低。
相反，5 分問兩題，分四次問，大把人爭住答～
你在點數計算上亦無吃虧～
原因好簡單：
20 分問七題，如果人地無咁多時間，或者只係識答其中三題，咁佢答唔答你好呢？
2014-02-22 21:11:09 補充：
我係意見欄 KO 左 第 7 題先啦～
#7
(a) Let f(x) = x²⁰⁰⁸ + 3, the required remainder is f(-1) = 1 + 3 = 4
2014-02-22 21:11:31 補充：
(b) From part (a), you know that when 6²⁰⁰⁸ + 3 is divided by 7, the remainder is 4.
If today is Saturday, then after (6²⁰⁰⁸ + 3 - 4) days, it is still Saturday;
2014-02-22 21:11:50 補充：
but after (6²⁰⁰⁸ + 3) days, that would be 4 days after Saturday, that is Wednesday.
2014-02-22 22:24:14 補充：
你咁好禮貌，我繼續答你吧．．．
3. If the roots of x² + ax + 12 = 0 are integers, what are the possible values of a?
Let the two roots be α and β.
{ α + β = -a
{ αβ = 12
The two roots are integers, we list all the factors of 12:
1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12
2014-02-22 22:25:38 補充：
The pair of (α, β) can only be the following (or reverse x and y)
(1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)
Therefore α + β can only be
13, 8, 7, -13, -8, -7
a can only be
-13, -8, -7, 13, 8, 7
2014-02-22 22:27:36 補充：
5.
If one root of the equation x² - (2k-5)x + k² - 5k + 56/9 = 0, where k is a constant, is twice the other root.
(a) Let the two roots be α and 2α.
α + 2α = 3α = 2k-5
Therefore, one root is α = (2k-5)/3
2014-02-22 22:29:42 補充：
(b) From (a), one root is (2k-5)/3, therefore, if we put x = (2k-5)/3, the equation would be satisfied. ＜－－ 重要 concept！！！
[(2k-5)/3]² - (2k-5)[(2k-5)/3] + k² - 5k + 56/9 = 0
(2k-5)²/9 - (2k-5)²/3 + k² - 5k + 56/9 = 0
-2(2k-5)²/9 + k² - 5k + 56/9 = 0
-2(2k-5)² + 9k² - 45k + 56 = 0
2014-02-22 22:30:37 補充：
-2(4k² - 20k + 25) + 9k² - 45k + 56 = 0
-8k² + 40k - 50 + 9k² - 45k + 56 = 0
k² - 5k + 6 = 0
(k - 2)(k - 3) = 0
k = 2 or k = 3
20 分 問 七題，有人答你的機會很低。
相反，5 分問兩題，分四次問，大把人爭住答～
你在點數計算上亦無吃虧～
原因好簡單：
20 分問七題，如果人地無咁多時間，或者只係識答其中三題，咁佢答唔答你好呢？
2014-02-22 21:11:09 補充：
我係意見欄 KO 左 第 7 題先啦～
#7
(a) Let f(x) = x²⁰⁰⁸ + 3, the required remainder is f(-1) = 1 + 3 = 4
2014-02-22 21:11:31 補充：
(b) From part (a), you know that when 6²⁰⁰⁸ + 3 is divided by 7, the remainder is 4.
If today is Saturday, then after (6²⁰⁰⁸ + 3 - 4) days, it is still Saturday;
2014-02-22 21:11:50 補充：
but after (6²⁰⁰⁸ + 3) days, that would be 4 days after Saturday, that is Wednesday.
2014-02-22 22:24:14 補充：
你咁好禮貌，我繼續答你吧．．．
3. If the roots of x² + ax + 12 = 0 are integers, what are the possible values of a?
Let the two roots be α and β.
{ α + β = -a
{ αβ = 12
The two roots are integers, we list all the factors of 12:
1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12
2014-02-22 22:25:38 補充：
The pair of (α, β) can only be the following (or reverse x and y)
(1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)
Therefore α + β can only be
13, 8, 7, -13, -8, -7
a can only be
-13, -8, -7, 13, 8, 7
2014-02-22 22:27:36 補充：
5.
If one root of the equation x² - (2k-5)x + k² - 5k + 56/9 = 0, where k is a constant, is twice the other root.
(a) Let the two roots be α and 2α.
α + 2α = 3α = 2k-5
Therefore, one root is α = (2k-5)/3
2014-02-22 22:29:42 補充：
(b) From (a), one root is (2k-5)/3, therefore, if we put x = (2k-5)/3, the equation would be satisfied. ＜－－ 重要 concept！！！
[(2k-5)/3]² - (2k-5)[(2k-5)/3] + k² - 5k + 56/9 = 0
(2k-5)²/9 - (2k-5)²/3 + k² - 5k + 56/9 = 0
-2(2k-5)²/9 + k² - 5k + 56/9 = 0
-2(2k-5)² + 9k² - 45k + 56 = 0
2014-02-22 22:30:37 補充：
-2(4k² - 20k + 25) + 9k² - 45k + 56 = 0
-8k² + 40k - 50 + 9k² - 45k + 56 = 0
k² - 5k + 6 = 0
(k - 2)(k - 3) = 0
k = 2 or k = 3
收錄日期: 2021-04-13 20:12:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140222000051KK00190