username1333513
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can anyone explain how to do these questions pleaseImage
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TenOfThem
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What have you tried so far

Do you know how to rationalise a denominator
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username1333513
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(Original post by TenOfThem)
What have you tried so far

Do you know how to rationalise a denominator
yes, i did that for the question right at the top, but i'm not getting my answer in the form they want it in, my calculator also had the same answer i got when i worked it out, the second i simply made a common denominator, did the workings out and got an answer of zero, the last one i have no idea how to do
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username1333513
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bump!!
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TenOfThem
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(Original post by bubblegumcat)
yes, i did that for the question right at the top, but i'm not getting my answer in the form they want it in, my calculator also had the same answer i got when i worked it out, the second i simply made a common denominator, did the workings out and got an answer of zero, the last one i have no idea how to do
So

When you rationalised the denominators for the 2 fractions in Q5 what did you get?
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WhiteGroupMaths
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(Original post by bubblegumcat)
yes, i did that for the question right at the top, but i'm not getting my answer in the form they want it in, my calculator also had the same answer i got when i worked it out, the second i simply made a common denominator, did the workings out and got an answer of zero, the last one i have no idea how to do
Q4(b): Did you multiply both numerator and denominator by 4- √7 ?

Q5: You shouldn't merge both blocks to achieve a common denominator in the first place. It is wiser to rationalize the two blocks separately.

Peace.
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Exon
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For Q4, recall the difference of two squares a^2 - b^2 = (a+b)(a-b) and use that to rationalise the denominator, using the surd as the b term.

For Q5, do as the above poster said. Rationalise first.

EDIT: Typo fixed
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Exon
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For 6, recall the formulae for the area of a sector and for the area of a triangle.

Area of a sector = \dfrac{\theta \pi r^2}{360}, where theta is the angle of the sector in degrees (hint: remember that the triangle is equilateral)

Area of a triangle = \dfrac{lh}{2}, where l = length, h = height.

Work out the height of the triangle by drawing a line from the midpoint of B and C (M) to A to make a right angle triangle. The size of line BM is half of the size of line BC, which is known (you know the radius and the triangle is equilateral). The size of line BA is also known (it is the radius). Use these two lines to work out the length of the third using pythagoras' theorem. You can now work out the area of the triangle.

Once you have both areas, you should know what to do.

EDIT: The Latex is a bit faulty. Fixing it.
EDIT2: Fixed.
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username1333513
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(Original post by TenOfThem)
So

When you rationalised the denominators for the 2 fractions in Q5 what did you get?
i rationalised and ended up with 2-2 =0, but thats not in the form they want it
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(Original post by Exon)
For Q4, recall the difference of two squares a^2 - b^2 = (a+b)(a-b) and use that to rationalise the denominator, using the surd as the b term.

For Q5, do as the above poster said. Rationalise first.

EDIT: Typo fixed
thanks, the explanation for q6 was very helpful! however for q4, i got (1+2√7)/9, im not sure if thats correct but its the only answer i can come up with
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username1333513
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(Original post by WhiteGroupMaths)
Q4(b): Did you multiply both numerator and denominator by 4- √7 ?

Q5: You shouldn't merge both blocks to achieve a common denominator in the first place. It is wiser to rationalize the two blocks separately.

Peace.
for q4 yes
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TenOfThem
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(Original post by bubblegumcat)
i rationalised and ended up with 2-2 =0, but thats not in the form they want it
Neither of these fractions rationalises to 2

The first fraction rationalises to \sqrt6 + \sqrt2
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(Original post by TenOfThem)
Neither of these fractions rationalises to 2

The first fraction rationalises to \sqrt6 + \sqrt2
oh right thanks, do you mind explaining how you rationalised the first fraction?
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TenOfThem
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(Original post by bubblegumcat)
oh right thanks, do you mind explaining how you rationalised the first fraction?
I multiplied the numerator and denominator by \sqrt3 + 1
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username1333513
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(Original post by TenOfThem)
I multiplied the numerator and denominator by \sqrt3 + 1
thank you!
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