The MEGA 9-1 GCSE Maths Exam Question Countdown - 30 DAYS TO GO
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Original post by Darwinion
Hmmm... I seem to be struggling with #69.
Area of triangle ABC = half of (x+3)(x+4)
In triangle DEF the missing angle is 30 degrees. And the area is therefore: 0.5 x (3x) x (x-1) x Sin 30
So I set up an equation as follows:
half of (x+3)(x+4) = half of (3x) (x-1) x Sin 30
Multiplying both sides by 2 gets rid of the fractions and Sin 30 = 0.5 so we now have:
(x+3)(x+4) = half of 3x (x-1)
Multiplying out the brackets I get:
x^2 + 7x + 12 = half of 3x^2 - 3x
Multiply both sides by 2 I get:
2x^2 + 14x + 24 = 3x^2 - 3x
So bringing over the terms from the left I now get:
x^2 - 17x - 24 = 0 and that's wrong :/
Area of triangle ABC = half of (x+3)(x+4)
In triangle DEF the missing angle is 30 degrees. And the area is therefore: 0.5 x (3x) x (x-1) x Sin 30
So I set up an equation as follows:
half of (x+3)(x+4) = half of (3x) (x-1) x Sin 30
Multiplying both sides by 2 gets rid of the fractions and Sin 30 = 0.5 so we now have:
(x+3)(x+4) = half of 3x (x-1)
Multiplying out the brackets I get:
x^2 + 7x + 12 = half of 3x^2 - 3x
Multiply both sides by 2 I get:
2x^2 + 14x + 24 = 3x^2 - 3x
So bringing over the terms from the left I now get:
x^2 - 17x - 24 = 0 and that's wrong :/
Try the question now
We forgot a factor of 1/2 when working it out initially
Original post by Darwinion
Spoiler
One day I'll log on to TSR fast enough to solve it before you.
Spoiler
I didn't see #68 till just now :/
(a) This simplifies to x^2 - x - 6 which factorises to: (x-3)(x+2)
(b)
(i) x^2 (x+2)
(ii) 4 (x+2)
(c) Using answers from (b) I can write: x^2 (x+2) - 4 (x+2) and so taking the common factor of (x+2) out I get: (x+2)(x^2-4)
Final check can be made by multiplying out the brackets to get back to the original expression in the question.
(a) This simplifies to x^2 - x - 6 which factorises to: (x-3)(x+2)
(b)
(i) x^2 (x+2)
(ii) 4 (x+2)
(c) Using answers from (b) I can write: x^2 (x+2) - 4 (x+2) and so taking the common factor of (x+2) out I get: (x+2)(x^2-4)
Final check can be made by multiplying out the brackets to get back to the original expression in the question.
(edited 7 years ago)
Original post by Darwinion
I didn't see #68 till just now :/
(a) This simplifies to x^2 - x - 6 which factorises to: (x-3)(x+2)
(b)
(i) x^2 (x+2)
(ii) 4 (x+2)
(c) Using answers from (b) I can write: x^2 (x+2) - 4 (x+2) and so taking the common factor of (x+2) out I get: (x+2)(x^2-4)
Final check can be made by multiplying out the brackets to get back to the original expression in the question.
(a) This simplifies to x^2 - x - 6 which factorises to: (x-3)(x+2)
(b)
(i) x^2 (x+2)
(ii) 4 (x+2)
(c) Using answers from (b) I can write: x^2 (x+2) - 4 (x+2) and so taking the common factor of (x+2) out I get: (x+2)(x^2-4)
Final check can be made by multiplying out the brackets to get back to the original expression in the question.
Is that fully factorised? What about ? Can it be reduced further?
Original post by crashMATHS
Is that fully factorised? What about ? Can it be reduced further?
Oh yeah... difference of two squares! (x+2)(x-2)
I always miss that. :/
here's yesterday's question
we're a bit caught up with an issue at the moment, so the questions in the countdown are likely to be a bit delayed for the next few days - we apologise!
we're a bit caught up with an issue at the moment, so the questions in the countdown are likely to be a bit delayed for the next few days - we apologise!
#70
(a) x^2 + y^2 = r^2 therefore equation: x^2 + r^2 = 25
(b) Subbing in the co-ordinates of the point (3,4) I get:
3^2 + 4^2 = 25 therefore 9 + 16 = 25 which is true and confirms that point (3,4) lies on the circle.
(c) The equation of the radius line from the origin (center of circle) to point (3,4) is: y= 4/3x
Equation of straight line is in the form y = mx + c where m = gradient and c = point the line crosses the y axis.
In this question the centre of the circle is at origin so c = 0
The tangent to the point (3,4) is perpendicular to the radius. Therefore the equation of this line is the negative reciprocal of the radius.
Therefore y = -3/4x for the tangent.
I think so anyway.
(a) x^2 + y^2 = r^2 therefore equation: x^2 + r^2 = 25
(b) Subbing in the co-ordinates of the point (3,4) I get:
3^2 + 4^2 = 25 therefore 9 + 16 = 25 which is true and confirms that point (3,4) lies on the circle.
(c) The equation of the radius line from the origin (center of circle) to point (3,4) is: y= 4/3x
Equation of straight line is in the form y = mx + c where m = gradient and c = point the line crosses the y axis.
In this question the centre of the circle is at origin so c = 0
The tangent to the point (3,4) is perpendicular to the radius. Therefore the equation of this line is the negative reciprocal of the radius.
Therefore y = -3/4x for the tangent.
I think so anyway.
(edited 6 years ago)
Original post by Darwinion
#70
(a) x^2 + y^2 = r^2 therefore equation: x^2 + r^2 = 25
(b) Subbing in the co-ordinates of the point (3,4) I get:
3^2 + 4^2 = 25 therefore 9 + 16 = 25 which is true and confirms that point (3,4) lies on the circle.
(c) The equation of the radius line from the origin (center of circle) to point (3,4) is: y= 4/3x
Equation of straight line is in the form y = mx + c where m = gradient and c = point the line crosses the y axis. In this question the centre of the circle is at origin so c = 0
The tangent to the point (3,4) is perpendicular to the radius. Therefore the equation of this line is the negative reciprocal of the radius.
Therefore y = -3/4x for the tangent.
I think so anyway.
(a) x^2 + y^2 = r^2 therefore equation: x^2 + r^2 = 25
(b) Subbing in the co-ordinates of the point (3,4) I get:
3^2 + 4^2 = 25 therefore 9 + 16 = 25 which is true and confirms that point (3,4) lies on the circle.
(c) The equation of the radius line from the origin (center of circle) to point (3,4) is: y= 4/3x
Equation of straight line is in the form y = mx + c where m = gradient and c = point the line crosses the y axis. In this question the centre of the circle is at origin so c = 0
The tangent to the point (3,4) is perpendicular to the radius. Therefore the equation of this line is the negative reciprocal of the radius.
Therefore y = -3/4x for the tangent.
I think so anyway.
Does the point , lie on the line ?
Original post by _gcx
Does the point , lie on the line ?
Ahhh... I jumped the gun at the end didn't I? I rushed and just did -ve reciprocal of the equation without realising that the y intersection is not zero for that line.
So... it should be (if I'm right)... y = -3/4x + 6.25 for the tangent
Thanks for that.
Original post by Darwinion
Ahhh... I jumped the gun at the end didn't I? I rushed and just did -ve reciprocal of the equation without realising that the y intersection is not zero for that line.
So... it should be (if I'm right)... y = -3/4x + 6.25 for the tangent
Thanks for that.
So... it should be (if I'm right)... y = -3/4x + 6.25 for the tangent
Thanks for that.
Yes, that is correct
WHY IS IT SO HARD??? 
Original post by ShiawaseNekox3
WHY IS IT SO HARD???
??
Are there any questions on here that you're struggling with, that you'd like me, or someone else, to you through?Original post by _gcx
?? Are there any questions on here that you're struggling with, that you'd like me, or someone else, to you through?
Are there any questions on here that you're struggling with, that you'd like me, or someone else, to you through?Do you get all the grade 8/9 stuff?
Original post by ShiawaseNekox3
Do you get all the grade 8/9 stuff?
Yes? Is there anything that you want me to explain?
Original post by _gcx
Yes? Is there anything that you want me to explain?
Everything please, i hate all of it
pm meQuick Reply
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