grizzleddude said:
Hey guys.
I was just wondering how to do these questions? I was away when my class did them so yeah I have little clue about the graphing etc. I know the gradient = rise over run but i don't know how to put it into practice...
http://img263.imageshack.us/img263/9742/mathsquestionsdh4.jpg
I also added in a few algebra ones which I find difficult... could anyone give me a hand figuring it out? I have always had trouble with damn algebra... haven't done it since term 1 which doesn't help either.
Thanks for any help!
Hello
I'll try my best to explain though i must warn you i am quite a horrible explainier =X
anyhow.
Question (ii) - graph y = 3x + 1
find x-intercept (where y = 0)
find y-intercept (where x = 0)
and just plot the points and draw the line
you should get x = -1/3
and y = 1
Question 16 -
Formular for a straight line equation is y=mx + b
m = gradient therefore like you said = rise/run
First thing you do is find the gradient. The gradient is rise/ run right?
Rise = y axis
Run = x axis
On the graph rise = 4 and run = 2
Therefore gradient = 4/2 = 2
Oh and also you notice that the line is sloping downwards therefore it has a negative gradient.
Therefore gradient = -2
And the b is the number where the line crosses the y-axis and in this case it is 4.
So with all this information we have we come up with the equation y=-2x+4
And so the answer would be C. It is just rearranged to y=4-2x which is the same thing as y=-2x +4
Question 23 -
y= 1/2x + 1
with this equation we know m=1/2 and that it is a positive slope.
and b= 1 therefore it crosses the y axis and 1
with this we can see that A is the answer.
Question (a) (i) -
just expand it
4p^(3)q^(4) (i.e. 4p to the power of 3, q to the power of 4 - i will be writing powers with ^ and the power number in brackets ( )
anyway i'll expand that part now
4p^(3)q^(4) x (-pq^2)
4p^(3) x -p = -4p^4
q^(4) x q^(2) = q^6 ====> Note: When multiplyig powers you add the powers.
therefore the answer is -4p^(4)q^(6)
I know the way i've written it seems confusing but i hope you get it anyway, i tried my best.
(ii)45xy^(2) divide by 15x^(2)
just put it over each other like this ===> 45xy^(2)/15x^(2)
and start cancelling,
so 45/15 = 3
you can cancel the x on the numerator (number at the top) and leave the y^(2) because there is no y in the numerator therefore you cannot cancel. Then, on the denominator you would have x instead of x^(2) because we got rid of the x at the top and so we have to get rid of 1 x at the bottom hence leaves us with just x on the denominator.
Hence, answer is 3y^(2) / x
Question (b) (i) - 5d(3a^(2) + ad)
first you get 5d and multiply it with 3a^(2) which = 15a^(2)d
second you get 5d and multiply with with ad which = 5ad^(2)
hence the answer = 15a^(2)d + 5ad^(2)
(ii) x(x-y) -y(y-x)
so once again expand. lets start wit x(x-y) with this you get x^(2)-xy
then do this with -y(y-x) and you get -y^(2)+xy (remember - x - = +)
so altogether we have x^(2) - xy - y^(2) + xy
now collect like terms.
hence, answer is =
x^(2) - y^(2)
Question (c) (i) 3t + 5 = 12 - t
collect like terms
3t + 5 = 12 - t ( i just moved the t over and remember when you move it over to the other side it becomes opposite ie. + becomes - and vice versa.
now add 3t + t which = 4t
and move the 5 over to the otherside and subtract it to 12 so,
12-5 = 7
now you have 4t = 7
make t the subject and you then have t = 7/4 which is the same as 1 3/4
Question (i) - im sure you can do this one just write down the y intercept (where the line crosses the y axis) and the gradient you can use rise/run or you can get the co-ordinates of x and y and use m = y2-y1/x2-x1 [im not sure if general uses this but if not just use rise over run
]
(ii) use y = mx + b remember b = y intercept and m = gradient
Well i hope this helps ! ehh im a bit late but i surely did take my time typing this up due to distractions and what not hehe
