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Maths year 11

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Original post by z_o_e


Oh I think I see where you're going wrong. It's just the way you write it.

You wrote 271327^{\frac{1}{3}} as 3273\sqrt{27} when you really meant it as 273\sqrt[3]{27}. Place the number in that little gap of the root as opposed to next to it, it's where I got confused.

Other than that, you got C and D wrong.
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Original post by RDKGames
Oh I think I see where you're going wrong. It's just the way you write it.

You wrote 271327^{\frac{1}{3}} as 3273\sqrt{27} when you really meant it as 273\sqrt[3]{27}. Place the number in that little gap of the root as opposed to next to it, it's where I got confused.

Other than that, you got C and D wrong.


How do I do c and D ?

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Original post by z_o_e
How do I do c and D ?

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For C, you were right at 1361/2\frac{1}{36^{1/2}} but you are wrong to say that this is the same as 36\sqrt{36}.
1361/2=136\displaystyle \frac{1}{36^{1/2}}=\frac{1}{\sqrt{36}}

For D, it's a similar problem. 1321/5=1325\displaystyle \frac{1}{32^{1/5}}=\frac{1}{\sqrt[5]{32}}. The fifth root of 32 is 2.
(edited 7 years ago)
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Original post by RDKGames
For C, you were right at 1361/2\frac{1}{36^{1/2}} but you are wrong to say that this is the same as 36\sqrt{36}.
1361/2=136\displaystyle \frac{1}{36^{1/2}}=\frac{1}{\sqrt{36}}

For D, it's a similar problem. 1321/5=1325\displaystyle \frac{1}{32^{1/5}}=\frac{1}{\sqrt[5]{32}}. The fifth root of 32 is 2.


How do I do b?


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Original post by z_o_e


You are correct at 1925\displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by 2525\frac{25}{25}. Also your C is wrong on that too.
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Original post by RDKGames
You are correct at 1925\displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by 2525\frac{25}{25}. Also your C is wrong on that too.


I got 225/25

So do I divide 25 by 225 and see what whole number goes into it?


What's wrong with C?


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Original post by RDKGames
You are correct at 1925\displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by 2525\frac{25}{25}. Also your C is wrong on that too.




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Original post by z_o_e


C is now fine.

For B:
1925⋅2525=1⋅25925⋅25=259\displaystyle \frac{1}{\frac{9}{25}} \cdot \frac{25}{25} = \frac{1 \cdot 25}{\frac{9}{25} \cdot 25}=\frac{25}{9}
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Original post by RDKGames
C is now fine.

For B:
1925⋅2525=1⋅25925⋅25=259\displaystyle \frac{1}{\frac{9}{25}} \cdot \frac{25}{25} = \frac{1 \cdot 25}{\frac{9}{25} \cdot 25}=\frac{25}{9}


How will I do these type of questions


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Original post by z_o_e
How will I do these type of questions


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I'll do the first example for you.

(8125)−4/3=(831253)−4=(25)−4=(2454)−1=(16625)−1=62516\displaystyle (\frac{8}{125})^{-4/3}=(\frac{\sqrt[3]8}{\sqrt[3]{125}})^{-4}=(\frac{2}{5})^{-4}=(\frac{2^4}{5^4})^{-1}=(\frac{16}{625})^{-1}=\frac{625}{16}
(edited 7 years ago)
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Original post by RDKGames
I'll do the first example for you.

(8125)−4/3=(831253)−4=(25)−4=(2454)−1=(16625)−1=62516\displaystyle (\frac{8}{125})^{-4/3}=(\frac{\sqrt[3]8}{\sqrt[3]{125}})^{-4}=(\frac{2}{5})^{-4}=(\frac{2^4}{5^4})^{-1}=(\frac{16}{625})^{-1}=\frac{625}{16}


I gave it a go ;//


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Original post by z_o_e
I gave it a go ;//


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Absolutely correct! :smile:
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Original post by RDKGames
Absolutely correct! :smile:


Yuss♥

Wb this?
Do I flip them over in the end?

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Original post by z_o_e
Yuss♥

Wb this?
Do I flip them over in the end?

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That's correct.

When you have a fraction raised to -1, the "flipping over" procedure is called taking the reciprocal of the fraction. It is easy to remember this, and well worth it.

The way it works is this;

Take (23)−1 \displaystyle (\frac{2}{3})^{-1}

This is the same as 123\displaystyle \frac{1}{\frac{2}{3}}

We can multiply the denominator by 3, which forces us to multiply the numerator by 3 as well:

123⋅33=1⋅323⋅3=32\displaystyle \frac{1}{\frac{2}{3}} \cdot \frac{3}{3} = \frac{1\cdot 3}{\frac{2}{3} \cdot 3} = \frac{3}{2} which is "flipped"
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Original post by RDKGames
That's correct.

When you have a fraction raised to -1, the "flipping over" procedure is called taking the reciprocal of the fraction. It is easy to remember this, and well worth it.

The way it works is this;

Take (23)−1 \displaystyle (\frac{2}{3})^{-1}

This is the same as 123\displaystyle \frac{1}{\frac{2}{3}}

We can multiply the denominator by 3, which forces us to multiply the numerator by 3 as well:

123⋅33=1⋅323⋅3=32\displaystyle \frac{1}{\frac{2}{3}} \cdot \frac{3}{3} = \frac{1\cdot 3}{\frac{2}{3} \cdot 3} = \frac{3}{2} which is "flipped"




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Original post by z_o_e


Correct.
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Original post by RDKGames
Correct.


How do I do this?


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Original post by z_o_e
How do I do this?


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You can treat the 32 as a fraction of 32/1 if you wish.

Otherwise:

(32)−4/5=(325)−4=(2)−4=(24)−1=(16)−1=116(32)^{-4/5}=(\sqrt[5]{32})^{-4}=(2)^{-4}=(2^4)^{-1}=(16)^{-1}=\frac{1}{16}
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Original post by RDKGames
You can treat the 32 as a fraction of 32/1 if you wish.

Otherwise:

(32)−4/5=(325)−4=(2)−4=(24)−1=(16)−1=116(32)^{-4/5}=(\sqrt[5]{32})^{-4}=(2)^{-4}=(2^4)^{-1}=(16)^{-1}=\frac{1}{16}


So for this... idk this is Confusing



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Original post by z_o_e
So for this... idk this is Confusing



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Yeah that's right. The cube root of 27 is....

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