# Nooo... Intermediate Maths QuestionsWatch

This discussion is closed.
Thread starter 13 years ago
#1
Yeah OK having trouble scraping a B here, maths being my weakest subject.. but anyway I have two questions here where I got a grand total of 0 marks on both. Can someone help me - HOW do you do these questions?! :

Question 1
Question 2
0
13 years ago
#2
For what value of x does the line cross the y-axis?
For what value of y does the line cross the x-axis?
Gradient = change in y/change in x- use the coordinates calculated in a)
0
13 years ago
#3
1)

rearrange to nice form:

2y - x = 4
2y = x + 4
y = 0.5x + 2

so, y = 0 when x = -4
x = 0 when y = 2

(by plugging the numbers in)

gradient, uh, well, by converting the equation into the above format we know the grad is a half. or, you can take the two points you know, (-4,0) and (0,2), and then find change in y over change in x.
0
13 years ago
#4
2)

angles in same segment... so it's 40 degrees.

opp. angles in cyclic quad... so it's 180 - 40 = 140 degrees.

since it's a tangent, BAD & BCD are right angles. ABCD is a quadrilateral -> internal angles = 360. 360 - 90 - 90 - 2(40) = 100
0
13 years ago
#5
Question 2
a) i) Angles in a segment rule. Look at the angle already drawn in. You don't have to do any calculations as it says 'write down...'

ii) Look at the quadrilateral with angles x and y in. There is a rule for quadrilaterals in a circle. As this is a one mark question, you don't have to do very much working out at all, just thinking.

b) What's angle ABD? Look at the angle you already know, and the fact that line AD is a tangent. Now you should know what CBD is. Add them together...
0
13 years ago
#6
Chewwy got there first! Sorry
0
13 years ago
#7
(Original post by hearthethunder)
Yeah OK having trouble scraping a B here, maths being my weakest subject.. but anyway I have two questions here where I got a grand total of 0 marks on both. Can someone help me - HOW do you do these questions?! :

Question 1
Question 2

Just for the sake of reiteration of what everyone else has said:

Question 1

2y - x = 4

Point A is on the x-axis, ie at a point where y=0.
Substitute y=0.
2(0) - x = 4,
-x = 4,
x=-4.
A(-4,0).

Similarly, Point B is on the y-axis, ie at a point where x=0.
Substitute x=0.
2y - (0) = 4
2y = 4
y = 2
So B(0, 2).

In a linear equation, that is, one in the form of y=mx + c, the gradient is m, and can be read off.

So we desire the given equation of the line, in this form - a simple rearrangement.

Given 2y - x = 4
2y = x + 4
y = 1/2(x + 4)
y = 0.5x + 2

So, the gradient m = 0.5.
0
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