jigoku_snow wrote: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.
Use the Distance Formula
to find the lengths of each base. Find the slope
of the line containing either base; use this to find the perpendicular slope. Find the line equation
for the perpendicular through one of the corners; find the point where this line intersects the other base (or the other base, extended). Find the distance between the two bases, and thus the height.
Plug the base lengths and the height into the area formula.
jigoku_snow wrote: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?
The line AB will be linear; what you have posted is a quadratic. How did you derive this? (One of the links in the previous response explains how to find linear equations from two points.)
As for finding the coordinates, try drawing the two points you have now, along with the line containing them. The rectangle is either to the one side of that line, or else to the other. So where must the other two points be?
jigoku_snow wrote: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?
Since nothing in the exercise says anything about ordering, combinations should be sufficient. However, you can see your book and your class notes; if your book, by default, defines "drawing" as "one at a time, with order taken into account", then you should use permutations.
jigoku_snow wrote: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 colours
i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1?
You seem to be counting the ways to choose the balls, but not the ways in which you could then distribute them (in the chosen groupings) amongst the boxes.
jigoku_snow wrote:ii) i have no clue at all for this. isnt it suppose to be the same as the first one?
If you have boxes A, B, and C and if you are placing three red balls in them, then you have the first situation. However, if you have boxes A, B, and C and if you are placing a red, a green, and a yellow ball in them, then you have the second situation. In the first case, "Box A has a red ball" is the same as "Box A has a red ball", since the balls are indistinguishable. In the second case, "Box A has a red ball" is different from "Box A has a green ball".
jigoku_snow wrote: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?
How does "D" differ from "D"? How are they distinguishable?
jigoku_snow wrote:6) 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?
What was your logic for including "2
jigoku_snow wrote: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).
[i]S5 = 2 (S15 - S5)
a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)]
How did you arrive at this equation?
Using the formulation explained here
Multiply by (1 - r)/a to get:
Then solve the polynomial
for its rational root (which clearly does not apply, as a value for the common ratio, to this series). Solve the resulting quadratic for the other root, using the definition of "geometric progression" to pick the correct value.