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Maths year 11

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Reply 1420
Original post by RDKGames
Factor out the numerator as much as you can, then see how much the denominator can divide it.


I divide the top and bottom by the same number?

They mustn't give decimals right?

Is the final answer a whole number or fraction?

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Original post by z_o_e
I divide the top and bottom by the same number?

They mustn't give decimals right?

Is the final answer a whole number or fraction?

Posted from TSR Mobile


Yes you divide the top and bottom by the same number as long as the top and bottom remain whole numbers.

No decimals.

The final answer is still a fraction but a simpler one. Smaller numbers.
Reply 1422
Original post by RDKGames
Yes you divide the top and bottom by the same number as long as the top and bottom remain whole numbers.

No decimals.

The final answer is still a fraction but a simpler one. Smaller numbers.


5 goes into -356

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Original post by z_o_e
5 goes into -356

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I have bad tendency to come up with difficult questions, but a challenge is always good! But just remember your normal procedure with surds and you'll get it right. Spoilers are there for hints. :smile:

A right-angled triangle has side length's (2+5)(2+\sqrt{5}) and (925)(9-2\sqrt{5})

(i) Find the length of the hypotenuse of this triangle and express it in the form ab5\sqrt{a-b\sqrt{5}} where aa and bb are integers. BONUS: Express the length of the hypotenuse in the form c55c45\sqrt{c} \sqrt{55-c^4\sqrt{5}} where c is an integer.

Spoiler

(ii) Express 4.1% of the triangle's area in the form AB(c3+55)\frac{A}{B}(c^3+5\sqrt5) where A, B and c are integers.

Spoiler

(iii) What is 1.4% of the area you found in part (ii)? Express this in the form (D102c)(c3+55)(D \cdot 10^{-2c})(c^3+5\sqrt5) where D is a decimal, and c is an integers.

Spoiler

(edited 7 years ago)
Reply 1424
Original post by RDKGames
I have bad tendency to come up with difficult questions, but a challenge is always good! But just remember your normal procedure with surds and you'll get it right. Spoilers are there for hints. :smile:

A right-angled triangle has side length's (2+5)(2+\sqrt{5}) and (925)(9-2\sqrt{5})

(i) Find the length of the hypotenuse of this triangle and express it in the form ab5\sqrt{a-b\sqrt{5}} where aa and bb are integers. BONUS: Express the length of the hypotenuse in the form c55c45\sqrt{c} \sqrt{55-c^4\sqrt{5}} where c is an integer.

Spoiler

(ii) Express 4.1% of the triangle's area in the form AB(c3+55)\frac{A}{B}(c^3+5\sqrt5) where A, B and c are integers.

Spoiler

(iii) What is 1.4% of the area you found in part (ii)? Express this in the form (D102c)(c3+55)(D \cdot 10^{-2c})(c^3+5\sqrt5) where D is a decimal, and c is an integers.

Spoiler



These are really hard :/

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Original post by z_o_e
These are really hard :/

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Don't be daunted by the looks of it, have a go and spend however much time you need to complete it. Ask whatever you need at each stage, you already know the methods to find what the questions asks for. This tests you on surds, algebra, fractions, and decimals altogether. :smile:
Reply 1426
Original post by RDKGames
Yes you divide the top and bottom by the same number as long as the top and bottom remain whole numbers.

No decimals.

The final answer is still a fraction but a simpler one. Smaller numbers.


So I square the brackets


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Original post by z_o_e
So I square the brackets


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Yes. Follow that formula, you know what a and b are. Try not to rush it, and check for any errors. :smile:
Reply 1428
Original post by RDKGames
Yes. Follow that formula, you know what a and b are. Try not to rush it, and check for any errors. :smile:


I got this


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Don't give it in decimals. Just expand the brackets and collect like terms.
Reply 1430
Original post by RDKGames
Don't give it in decimals. Just expand the brackets and collect like terms.




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That's not what I meant.
c2=(2+5)2+(925)2\displaystyle c^2=(2+\sqrt5)^2+(9-2\sqrt5)^2

=(2+5)(2+5)+(925)(925)\displaystyle =(2+\sqrt5)(2+\sqrt5)+(9-2\sqrt5)(9-2\sqrt5)

THOSE brackets. First expand the two then add up any common terms.
(edited 7 years ago)
Reply 1432
Original post by RDKGames
That's not what I meant.
c2=(2+5)2+(925)2\displaystyle c^2=(2+\sqrt5)^2+(9-2\sqrt5)^2

=(2+5)(2+5)+(925)(925)\displaystyle =(2+\sqrt5)(2+\sqrt5)+(9-2\sqrt5)(9-2\sqrt5)

THOSE brackets. First expand the two then add up any common terms.


Like this?


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Yep you expanded that one correctly; so now you know that (2+5)2=9+45(2+\sqrt5)^2=9+4\sqrt5, now do the other one. :smile:
Reply 1434
Original post by RDKGames
Yep you expanded that one correctly; so now you know that (2+5)2=9+45(2+\sqrt5)^2=9+4\sqrt5, now do the other one. :smile:


Is this correct?


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Original post by z_o_e
Is this correct?


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First 3 terms are correct, but check that last one. (25)(25)(-2\sqrt5)(-2\sqrt5) is not 45-4\sqrt5
Reply 1436
Original post by RDKGames
First 3 terms are correct, but check that last one. (25)(25)(-2\sqrt5)(-2\sqrt5) is not 45-4\sqrt5


-20?

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Reply 1437
Original post by RDKGames
First 3 terms are correct, but check that last one. (25)(25)(-2\sqrt5)(-2\sqrt5) is not 45-4\sqrt5




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Correct number but wrong sign :smile: Notice that you are multiplying negative by a negative there so the answer must be a positive. Other than that, you have now expanded both brackets. Refer back to the Pythagorean theorem and see what you need to do with the two answers you just got.
Reply 1439
Original post by RDKGames
Correct number but wrong sign :smile: Notice that you are multiplying negative by a negative there so the answer must be a positive. Other than that, you have now expanded both brackets. Refer back to the Pythagorean theorem and see what you need to do with the two answers you just got.


How's this?


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