Mathematics Question Thread

[RuiAce]

Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)

For some reason I could not count. Or I interpreted 5 as 6.

My apologies. Yes, they (and I) are wrong. kemi's remark on the mistake in Excel Success One could prove useful.

[RuiAce]

Hey Guys!!

Sorry i'm quite stuck on a certain question! A textbook question mind u!!! 

Q13

The half-life of radium is 1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.

Thanks in advance!!

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[Justinhales]

Thankyou!!   

Makes sense now!!

[sidzeman]

Could someone please help me with part iii, iv and v

[RuiAce]

Could someone please help me with part iii, iv and v

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This is just the formula \(A = \frac{1}{2}r^2\theta\). Just remember to use the correct value for \(\theta\), i.e. the one for the major sector. Not the minor sector.
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Find the area of the major sector AOB using the same formula as above, and then just add all the areas you had from iii and iv with that one.

[sidzeman]

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This is just the formula \(A = \frac{1}{2}r^2\theta\). Just remember to use the correct value for \(\theta\), i.e. the one for the major sector. Not the minor sector.
________________________________

Find the area of the major sector AOB using the same formula as above, and then just add all the areas you had from iii and iv with that one.

For part iii, I proved the triangles were congruent as well, but then used 1/2absinC (times 2) to calculate the area, and got 2root3.
My working out was A= 2(0.5)(sin60)(2)(1) - using the absolute value for sin (root3 /2) - shouldnt the answer be 2root3 or am I doing something wrong?

For part iv - i used the angle pi/3 is this not correct?

[RuiAce]

For part iii, I proved the triangles were congruent as well, but then used 1/2absinC (times 2) to calculate the area, and got 2root3.
My working out was A= 2(0.5)(sin60)(2)(1) - using the absolute value for sin (root3 /2) - shouldnt the answer be 2root3 or am I doing something wrong?

For part iv - i used the angle pi/3 is this not correct?

Your expression "A= 2(0.5)(sin60)(2)(1)" already counts the two triangles together due to that extra '2' in front. Else, you fudged something in when you should not have.
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[clarence.harre]

\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}

There weren't any traps here.

Ok Great!! So I guess I didn't make any mistakes in the working but I still don't understand how if 5 + 1 = 6 but 5 + 5 = 11. I still swear it's 10; even plugged it into a time difference calculator (see attachtment). Thank you RuiAce for the prompt response though

[RuiAce]

Ok Great!! So I guess I didn't make any mistakes in the working but I still don't understand how if 5 + 1 = 6 but 5 + 5 = 11. I still swear it's 10; even plugged it into a time difference calculator (see attachtment). Thank you RuiAce for the prompt response though

I apologised about this one at the top of this page aha

[nattynatman]

this might be a stupid question...

but which questions require you to use the discriminant to prove it exists?
I always seem to miss using it mostly cos I'm not sure when to apply it

[RuiAce]

this might be a stupid question...

but which questions require you to use the discriminant to prove it exists?
I always seem to miss using it mostly cos I'm not sure when to apply it

When you're interested in the number of solutions, not the actual solution itself.

Sure, there are a whole bunch of applications of the discriminant, and they generally target band 6 as well. A common one can be in the number of points of intersections. But ultimately they still rely on the number of solutions to a quadratic equation you get given.

[K9810]

Hi, I was wondering if you can please explain why the first line of the answers for part iii) says angle CBD=36

[RuiAce]

Hi, I was wondering if you can please explain why the first line of the answers for part iii) says angle CBD=36

Because if you replicate the exact same proof for how you found \(\angle BAC\), but just replaced some of the sides, you would get the exact same answer for \(\angle CBD\).

"Similarly" can only be used when you try to find something else, but the steps in the proof goes the exact same way.

(Alternatively by inspection \( \triangle BAC \equiv \triangle CBD \) (SAS) )

[K9810]

Because if you replicate the exact same proof for how you found \(\angle BAC\), but just replaced some of the sides, you would get the exact same answer for \(\angle CBD\).

"Similarly" can only be used when you try to find something else, but the steps in the proof goes the exact same way.

(Alternatively by inspection \( \triangle BAC \equiv \triangle CBD \) (SAS) )

Thank you! That makes so much more sense

[asd987]

Hi can i get help with ii) please. thanks