Mathematics Question Thread

[Mohammad130no]

Hey! So, the equation we have been given is



and we want to rewrite it in standard form. First, we need to complete the square, as the final answer is in the form (x-h)^2. So,



And that's our answer! We got



Great! Now, we want to calculate the roots of the function. The easiest way of doing this is the quadratic equation.



Easy as Now, with all of the above information (and subbing in x=0, y=0 etc.) you can easily sketch the graph!

For the last part, we essentially want to know where f(x) is ABOVE the x-axis. We know the function is a positive parabola, so this will occur to either side of the roots (ie. to the left of the first x intercept, and to the right of the second x intercept). This will become super obvious once you draw the sketch

Cheers for all the help.

So thats it for b) ?? the roots are just 4+-(sqrt)30?

[RuiAce]

Cheers for all the help.

So thats it for b) ?? the roots are just 4+-(sqrt)30?

Yes

[Mohammad130no]

Yes

Thank you guys so much

[jakesilove]

Sorry jake I am having a bit of difficulty when graphing this function. If I find the x intercepts mathematically it is (4+-[sqrt30])/2 , and y int is -7. Though if I put it into a graphing website it shows the y int as (2,-15)

The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.

[Mohammad130no]

The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.

all graphing wesbites show that as the x value. And you were right about the y and vertex, i messed up

[kiwiberry]

all graphing wesbites show that as the x value. And you were right about the y and vertex, i messed up

silly mistake with the quadratic formula haha

[jakesilove]

The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.

Aha true, true, you can see which line I messed up above (forget the 2*a in the quadratic equation!). So you're all sorted now?

[jakesilove]

Help Please, this is from the Christian 2011 boys High school Trial Paper

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Could you post this up on the Physics thread?

[jakesilove]

Sure, even though its from the 2 unit paper?

About the statement you said before where you made a mistake, are you refering to b) where you found the roots?

Aha that's crazy, I'm clearly having a bad day sorry about that! Just assumed it was Physics (Keplar's law is in the Space curriculum, and never seen a Maths question on it!). I'll have a crack now, and yet I just made a mistake in finding the roots.

[jakesilove]

Help Please, this is from the Christian 2011 boys High school Trial Paper

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So, for the first part (proportional to), we would write



As a full equation, we can replace the 'proportional to' with some constant, k.



Now, we know that the Earth will rotate around the sun in one year. We also know that the average distance is 150 million kilometers. So, we can write




Okay, so we have



Part c) doesn't actually have a question; I assume it wants us to find the period of Neptune? Well, the distance is 30*150 million km.




Google says that the orbital period is about 164 years, so this approximation works well!

Now, for the final part, we again know that



The k may be different, who knows. However, we assume that the k value is the same for planet X and planet Y. Let the distance between planet X and the Sun be



Then,




Now, Y is twice the distance away, so





So,



Thus, planet Y has an orbital period that is about 35% greater than planet X

[JuliaPascale123]

Heeyyy, this question was ridiculous. Can I please have some help

[jakesilove]

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Heeyyy, this question was ridiculous. Can I please have some help

Hey Julia! Welcome to the forums!

The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,







etc.

Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be



Now, we know that the initial quantity is 300mg. So




We also know that the quantity has dropped to 225mg after 40 minutes. So,





So, the equation for the quantity of the medicine is



Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.





Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.

Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!



is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).

Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,




But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.

The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.

For the last part, we solve for




Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.

[itssona]

Heey,

with absolute value, I was doing problems and wanted to confirm;

if I have an inequality with x on both sides, do i square both sides??

like in
|x-2|<x

and how would you show this on a numberline:
|4x-2| >=3

thanks!!

[RuiAce]

Heey,

with absolute value, I was doing problems and wanted to confirm;

if I have an inequality with x on both sides, do i square both sides??

like in
|x-2|<x

and how would you show this on a numberline:
|4x-2| >=3

thanks!!

______________

[RuiAce]