sums ....plzz help?

Question

1) Solve

x^2+3x+2/x^2+5x+4 = 3/2

Where x not equal to -1 and x not equal to -4

2)

http://img411.imageshack.us/img411/2008/newbitmapimagews3.png

In the given figure AB=AC and angle ACD = 120
Find ‘angle’ A

3)
In the given figure AB=CF, EF=BD
‘angle’AFE = ‘angle’DBC
Prove that

‘triangle’AFE ‘Congruent to’ ‘triangle’CBD

http://img519.imageshack.us/img519/1885/newbitmapimage2pn0.png


4)
In the given figure X and Y are two points on equal sides AB and AC of ‘triangle’ABC such that AX=AY. Prove that XC=YB

http://img519.imageshack.us/img519/7133/newbitmapimageub1.png


5)
In the given figure ‘angle’A = ‘angle’C and AB=BC
Prove that ‘triangle’ABD ‘Congruent to’ ‘triangle’CBE


http://img293.imageshack.us/img293/6393/newbitmapimage2jz4.png


6)

In the given figure S is any point on the side QR of ‘triangle’PQR. Prove that PQ+QR+RP ’greater than’ 2PS

http://img339.imageshack.us/img339/9617/newbitmapimageoq2.png

FIRST QUESTION IS SUMTHING LIKE THIS:

x^2 + 3x +2 3
_________ = ___
x^2 + 5x +4 2

where x is not equal to -1
and x is not equal to -4

Answer 1

(1) I assume you mean the entire polynomials are divided. It would be better to supply parentheses:

(x^2+3x+2)/(x^2+5x+4) = 3/2
(x + 2)(x + 1) / (x + 4)(x + 1) = 3/2

Cancel out the x+1:

(x + 2)/(x + 4) = 3/2

Cross-multiply:

2(x + 2) = 3(x + 4)
2x + 4 = 3x + 12
x = -8

Check the answer:

(x^2+3x+2)/(x^2+5x+4) =
((-8)^2 + 3(-8) + 2)/((-8)^2 + 5(-8) + 4) =
(64 - 24 + 2) / (64 - 40 + 4) =
42 / 28 =
3*14 / 2*14 = 3/2

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(2) Angle ACB must be 60 because it is supplementary to the angle labeled 120.

Angle ABC must also be 60 because the triangle is isoceles -- it must be the same as angle ACB.

And, since the sum of three angles in a triangle are 180, that leaves 180 - 60 - 60 = 60 for angle A.

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(3) Since AB = CF,
AB + BF = CF + BF.
But AB+BF = AF, and CF+BF = CB, thus:
AF = CB

AF/CB is one side of angle AFE/DBC, and the other side of that angle (EF/BD) is already given as congruent.

Thus the two triangles are congruent by SAS.

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(4) AX=AY is given, and AB=AC is given. Then it is also true that AB-AX = AY-AC, i.e. XB = YC.

Since AB=AC the triangle is isoceles, and this means that angle ABC equals angle ACB.

Side BC is equal to side BC.

So in triangle XBC, you have XB, angle B, and BC congruent, respectively to YC, angle C, and BC of triangle YCB.

By SAS the two triangles are congruent.

And that means that the corresponding parts, sides XC and YB in particular, are also congruent.

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(5) You're given
(a) angles A and C are congruent
(b) AB is congruent to CB

And you know that angle B is congruent to itself.

Then by ASA (the side AB/CB and the angle on either side of it in triangles ABD and CBE respectively), triangles ABD and CBE are congruent.

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(6)
It's not a formal proof, but I think you can turn it into one:

Consider the longer of the two sides PQ and PR. Let's say that it is PQ (though it doesn't matter, and you can pick either if they are equal).

PS must be shorter than or equal to PQ. (It's only equal if S and Q are the same point.)

It is a rule that the sum of any two sides of a triangle must be greater than the third side. So we also know that PR + QR > PQ

Since PR + QR > PQ
We know PQ + PR + QR > 2PQ
(just added PQ to both sides)

And since PQ >= PS,
We know 2PQ >= 2PS
(multipled both sides by 2)

From those two (PQ + PR + QR > 2PQ, 2PQ >= 2PS), we know:
PQ + PR + QR > 2PS

Answer 2

1) x = -7.554098838, or x = -0.851876428, either one gives 2/3 for the equation.

But ... the equation look wrong. Why does it have 3x and 5x? They should just be added up to 8x. Did you leave out something?

Answer 3

(x^2+3x+2)/(x^2+5x+4) = 3/2
(x+2)(x+1)/(x+4)(x+1)=3/2
2x+4=3x+12
-x=8
x=-8


2. 60

i don't have to prove anything

Answer 4

this is for # 2.... angle A equals 60.

Answer 5

enable me provide a less complicated thank you to remedy the two equtions have been given by ability of all enable p=perimeter, l=length and b=breadth so2l+2b = 2 (l+b) = p =240 so l+b = 240/2 =one hundred twenty additionally 2( 0.9l+a million.2b) = 240 so 0.9l+a million.2b = one hundred twenty so l+b = 0.9l+a million.2b subtracting 0.9l and a million.2b from the two thesides 0.1l-0.2b= 0 including 0.2b to the two the aspects 0.1l = 0.2 b dividin the two aspects by ability of 0.a million l = 2b considering l+b =one hundred twenty 2b+b =one hundred twenty 3b = one hundred twenty dividing the two aspects by ability of three b = one hundred twenty/3 = forty considering l = 2b l = 2*forty = eighty so the dimensions and breadth of the rectangle is eighty cm. and forty cm. respectively (examine: while length of 80cm is decreased by ability of 10% it turns into seventy two cm. and breadth of forty cm. is greater by ability of 20% it turns into forty 8 cm. so the fringe turns into 2(seventy two+forty 8) = 240 cm. that's the comparable fee of the fringe with breadth of forty cm and length of 80cm)

Answer 6

what the heck
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