need help (1 Viewer)

[Mr Chi]

1) In a throw of two dice, find the probability of scoring more than 6.

2) ABC is a triangle in which B= 135 deg A= 30 deg and b = 10cm ; find a in surd form

3) Find the equation of a straight line which is perpendicular to 3x - 2y -7 = 0 and passes through ( 2, -1 )

any help in working these out ?, thanks

 

[Riviet]

1) In a throw of two dice, find the probability of scoring more than 6.

2) ABC is a triangle in which B= 135 deg A= 30 deg and b = 10cm ; find a in surd form

3) Find the equation of a straight line which is perpendicular to 3x - 2y -7 = 0 and passes through ( 2, -1 )

any help in working these out ?, thanks

1) Draw a table with 36 squares, excluding the columns and rows where you mark in the different data, leave the 36 squares unmarked and then tick the ones that add up to 7 through to 12 because these are the sums of the two dice rolls that are greater than 6. Now divide the number of ticks by 36 because that's the total number of possible outcomes that are possible and you have your answer.

2) Use the sine rule. Since you are finding a side, use a/sinA=b/sinB and substitute what you are given in the question. Don't forget the exact ratios e.g sin30^o=1/2.

3) Firstly find the gradient of the perpendicular line by either using m=-b/a OR express the equation 3x - 2y -7 = 0 in the form y=mx+b, then using the point gradient formula y-y₁=m(x-x₁) to find the equation of the line.

 

[YBK]

2) ABC is a triangle in which B= 135 deg A= 30 deg and b = 10cm ; find a in surd form

3) Find the equation of a straight line which is perpendicular to 3x - 2y -7 = 0 and passes through ( 2, -1 )

any help in working these out ?, thanks

umm.... as Riv said, sine rule for the first



For the second u can do this; it's a cool way of finding perpendiculars..

make it, 3x - 2y = 7
Now the rule is to swap the numbers around and change the sign
So it become 2x + 3y = n
Now replace the sub ( 2, -1 ) in equation
and u get 4 - 3 = 1
.: n = 1
.: Equation of perpendicular is 2x+3y = 1

 

[Riviet]

For the second u can do this; it's a cool way of finding perpendiculars..

make it, 3x - 2y = 7
Now the rule is to swap the numbers around and change the sign
So it become 2x + 3y = n
Now replace the sub ( 2, -1 ) in equation
and u get 4 - 3 = 1
.: n = 1
.: Equation of perpendicular is 2x+3y = 1

Interesting shortcut there, but I would only use that to check your answer quickly.

 

[YBK]

Interesting shortcut there, but I would only use that to check your answer quickly.

You could actually use that formally. There's a proper way to set it out, to get marks for working; i'll give another example.



Find the equation of the line through (2,1) which is perpendicular to 3x + 2y - 5 = 0


Solution:

Equation of perpendicular line is
2x - 3y = k for some constant k

(2,1) lies on the line
4-3 = k .: k = 1

Ans: 2x - 3y = 1
2x - 3y + 1 = 0



The advantages of this method are obvious.. and especially obvious for the parametrics chapter, where they often tell you to simply write the equation of the perpendicular (usually involving focal length), and it'd be a waste of time to put the whole thing in y-y1 = m (x - x1)

You should do a few examples first to become familiar with the method first though....

 

[insert-username]

That is a very nice, fast method, YBK. Me likey.


I_F