plz help me to solve the following questions as much as u can,thx in advance!
1.a piece of rod is cut into 2 at a pt selected at random.the probability that the longer part is at least 5 times as long as the shorter part is?
2.if (-2,4) is equidistant from the pts (-3,2) and (0,b),then b =?
3. Given 2 ts A (x1,y1) and B (x2,y2).If P is a pt on AB and AP:PB=r:s,then P=?
4. Plz help me to check whats wrong with my caluculation:
Let A=(-4,6),B=(4,2),C is a pt on BA produced such that AC = 1/5BC.The x-coordinate of the pt C is:
AC:BC=1:5
1(4) + 5(-4)/1+5 = -16/6 = -8/3
5.Find the values of A,B and C if 3+Ax+Bx^2 = Ax(x-1) + C(x-1) is an identity in x.
6. Given that x^2 – 5x +7 = (x+a)^2 + b for all x,find the values of the constants a and b.
7. If 3x+2/(x-1)^2(2x-3) = A/x-1 + B/(x-1)^2 + C/2x-3 , find the values of A,B and C.
maths problems.
2009-03-19 8:18 am
回答 (4)
2009-03-21 6:17 am
2009-03-25 5:23 am
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2009-03-19 10:14 pm
No.4
It should be
AC:BC = 1:5
CA:AB = 1:4
Let the x-coordinate of the pt C be x.
4x + 1(4) = -4
x = -2
No.8
Is there any mistake in the Qu?
No.9
When f(x) = 2x^3 + ax^2 + bx + c is divided by (x^2 - 1),
the remainder is (b+2)x + a+c
Hence (b+2)x + a+c = x+1
b+2 = 1 and
a+c = 1 ...... (1)
We have b = -1
Given that (x-2) is a factor of f(x),
f(2) = 0
2(2)^3 + a(2)^2 + -(2) + c = 0
4a + c = -14 ...... (2)
Solving (1) and (2)
We have a = -5 and c = 6
No.10
f(x) and g(x) have a common factor (x-p)
f(p) = g(p) = 0
f(p) - g(p) = 0
Therefore (x-p) is also a factor of the polynomial f(x) - g(x)
If ax^3 + 4x^2 - 5x - 10 = 0 and ax^2 -9x - 2 = 0 have a common root,
then ax^3 + 4x^2 - 5x - 10 = 0 and ax^3 -9x^2 - 2x = 0 also have a common root, say p.
By the above result,
ax^3 + 4x^2 - 5x - 10 - (ax^3 -9x^2 - 2x) = 0 also have a root p.
ap^3 + 4p^2 - 5p - 10 - (ap^3 -9p^2 - 2p) = 0
13p^2 - 3p -10 = 0
p = 1 or -10/13
Substitute p = 1 into ax^2 -9x - 2 = 0 and ax^3 + 4x^2 - 5x - 10 = 0, we have a = 11
Substitute p = -10/13 into ax^2 -9x - 2 = 0 and ax^3 + 4x^2 - 5x - 10 = 0, we have a = -8.32 (Why you get 2?)
It should be
AC:BC = 1:5
CA:AB = 1:4
Let the x-coordinate of the pt C be x.
4x + 1(4) = -4
x = -2
No.8
Is there any mistake in the Qu?
No.9
When f(x) = 2x^3 + ax^2 + bx + c is divided by (x^2 - 1),
the remainder is (b+2)x + a+c
Hence (b+2)x + a+c = x+1
b+2 = 1 and
a+c = 1 ...... (1)
We have b = -1
Given that (x-2) is a factor of f(x),
f(2) = 0
2(2)^3 + a(2)^2 + -(2) + c = 0
4a + c = -14 ...... (2)
Solving (1) and (2)
We have a = -5 and c = 6
No.10
f(x) and g(x) have a common factor (x-p)
f(p) = g(p) = 0
f(p) - g(p) = 0
Therefore (x-p) is also a factor of the polynomial f(x) - g(x)
If ax^3 + 4x^2 - 5x - 10 = 0 and ax^2 -9x - 2 = 0 have a common root,
then ax^3 + 4x^2 - 5x - 10 = 0 and ax^3 -9x^2 - 2x = 0 also have a common root, say p.
By the above result,
ax^3 + 4x^2 - 5x - 10 - (ax^3 -9x^2 - 2x) = 0 also have a root p.
ap^3 + 4p^2 - 5p - 10 - (ap^3 -9p^2 - 2p) = 0
13p^2 - 3p -10 = 0
p = 1 or -10/13
Substitute p = 1 into ax^2 -9x - 2 = 0 and ax^3 + 4x^2 - 5x - 10 = 0, we have a = 11
Substitute p = -10/13 into ax^2 -9x - 2 = 0 and ax^3 + 4x^2 - 5x - 10 = 0, we have a = -8.32 (Why you get 2?)
2009-03-19 7:15 pm
1. 1/6
2. [(-2+3)2+(4-2)2]=[(-2-0)2+(4-b)2]
b=3
3. P ((sx1+rx2)/(r+s) , (sy1+ry2)/(r+s) )
4. nothing wrong
5. 3+Ax+Bx2=Ax2+(C-A)x-C
C=-3
3+Ax+Bx2=Ax2(-3-A)x+3
A=-3-A
2A=-3 A=-3/2
B=A = -3/2
6. x2-5x+7=x2+2ax+(a2+b)
2a=-5 a=-5/2
a2+b=7 b=3/4
7. first of all, 將右邊通分母,then you will get
3x+2 = (2A+C)x2+(2B-5A-2C)x+(3A-3B+C)
2A+C = 0 2B-5A-2C=3 3A-3B+C=2
C=-2A A=2B-3 B=-5
C=-2[2(-5)-3]=26 A=-13
8. i dun no
9. also i dun no
2. [(-2+3)2+(4-2)2]=[(-2-0)2+(4-b)2]
b=3
3. P ((sx1+rx2)/(r+s) , (sy1+ry2)/(r+s) )
4. nothing wrong
5. 3+Ax+Bx2=Ax2+(C-A)x-C
C=-3
3+Ax+Bx2=Ax2(-3-A)x+3
A=-3-A
2A=-3 A=-3/2
B=A = -3/2
6. x2-5x+7=x2+2ax+(a2+b)
2a=-5 a=-5/2
a2+b=7 b=3/4
7. first of all, 將右邊通分母,then you will get
3x+2 = (2A+C)x2+(2B-5A-2C)x+(3A-3B+C)
2A+C = 0 2B-5A-2C=3 3A-3B+C=2
C=-2A A=2B-3 B=-5
C=-2[2(-5)-3]=26 A=-13
8. i dun no
9. also i dun no
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