**1) ABCD is a parallelogram and the coordinates of A, B, C are (-8, -11), (1, 2) and (4, 3) respectively. Find the area of the parallelogram.**

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*i manage to find that D =( -5.-10) . area of parallelogram = AD x height i'm stuck here. how to find the height? i tried it by finding the angle between AB and AD first , (37.32 degree) then use it to find the vertical line from B. but i dont get the right answer.***2). Points A and C have coordinates (-1, 2) and (9, 7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units.***from the question, i know that AB=CD=10. the equation of line AB = x^2 +y^2+ 2x -4y-97=0. the equation of line CD= a^2 +b^2-18 a- 14b +30=0. then what should i do next?***3)If the five letters a, b, c, d, e are put into a hat, in how many ways could you draw two letters?***must we use permutation or combination? if permutation, why? why can't i use combination? 5C2?*4))

**3 balls are to be placed in 3 different boxes, not necessarily with 1 ball in each box. any box can hold up to 3 balls. find the number of ways the balls can placed if**

i) they are all of the same colour

ii) they are all of different coloursi) they are all of the same colour

ii) they are all of different colours

*i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1?*

ii) i have no clue at all for this. isnt it suppose to be the same as the first one?ii) i have no clue at all for this. isnt it suppose to be the same as the first one?

**5)GOLD MEDAL- if 2 D letters come first and the 2 L letters come last, find how many different arrangement there are***D D _ _ _ _ _ L L*

5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L?5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L?

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**3 letters are selected at random from the word "BIOLOGY" . Find the number that the selection contains one letter O***2C1 x 5C2 = 20 , the answer given is 5C2 = 10 , why i dont have to write 2C1?*7)

**The sum of the first 5 terms of a G.P. is twice the sum of the terms from the 6th to the 15th inclusive. Show that r^5 = 1/2 (SQRT3 -1).***S5 = 2 (S15 - S5)*

a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)]

1 - r^5 = 2(r^5 -r^15)

2r^15-3r^5+1=0

let u=r^5

2u^3 - 3u +1=0

i'm stuck here. what to do next?a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)]

1 - r^5 = 2(r^5 -r^15)

2r^15-3r^5+1=0

let u=r^5

2u^3 - 3u +1=0

i'm stuck here. what to do next?