Mathematics Question Thread

[jamonwindeyer]

Hello
May I have help with this question? (Part a)
Excuse the scribbling thanks

Hey!! So we can deduce fairly easily that \(\triangle ATD\) and \(\triangle FTB\) are congruent (use an AAS proof), so you'd start there. This lets you prove that \(AT=FT\). Now, assign a variable to \(AT\), let it be \(x\), and label \(BT=7-x\) accordingly. From there, use pythagoras theorem in \(\triangle BFT\):

[jamonwindeyer]

Hey! can someone explain how to do this question??
I'm usually stumped when questions like these show up.

Hey! So when going from a derivative to an original function, remember that the given graph represents the gradient of the graph you are drawing. So, for this line, the gradient starts negative, and gradually increases to a positive value. The line turns into a parabola, that decreases initially (negative gradient), turns around where the line intercepts the axis, then increases from there. That's the red curve.

Doing the same with the red curve, we simply mark where the red curve is positive - Our blue graph slopes UPWARDS there. When the red curve is negative, our blue graph slopes DOWNWARDS. Again, the intercepts of the red curve are where the blue curve turns around!

Note also that \(f'(0)=a, a>0\) means the first derivative needs to have a y-intercept above the x-axis. My red curve does this (accidentally to be honest, I've only just read that bit )

[gilliesb18]

Hello-
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!

Thanks 😀😀

[Natasha.97]

Hello-
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!

Thanks 😀😀

Hi!

I've attached my solution to part g

Hope this helps

Edit: Attached the rest of the solutions 

[Shadowxo]

Hello-
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!

Thanks 😀😀

Let's say the angle inside the brackets is a*x (x is easier to type than theta)
First find a*x (like you'd do if it were just x by itself)
Then divide all the solutions by a to get x by itself

eg g)
Tan(2x) = sqrt(3)
2x = 60°, 240°, 420°, 600°
(Note 420 = 60+360, 600=240+360)
x = 30°, 120°, 210°, 300°

Also you should know the table of values off by heart (or the triangle method for memorising them).

[gilliesb18]

Hi!

I've attached my solution to part g (not enough time for the rest)

Hope this helps

Thanks sooooo sooo much!!!!! Very kind:)

[_____]

If they give us the rate of change (kP), can I just say P = P0e^kt or do I need to do some proof to get the population equation? No need to solve this for me I'm just wondering if I need to show an integration of kP and if so, how I'd do that.



Stuck with IV on this one, answers I've seen don't make much sense. I tried to set 1 - 2cosΘ >= 0 as a and y need to be positive but I wasn't able to get the answer.

Thanks.

[RuiAce]

(Image removed from quote.)

If they give us the rate of change (kP), can I just say P = P0e^kt or do I need to do some proof to get the population equation? No need to solve this for me I'm just wondering if I need to show an integration of kP and if so, how I'd do that.

(Image removed from quote.)

Stuck with IV on this one, answers I've seen don't make much sense. I tried to set 1 - 2cosΘ >= 0 as a and y need to be positive but I wasn't able to get the answer.

Thanks.

If you want to be safe you can prove it anyway. But personally I'd just assume it, because in 2U you aren't taught how to properly solve that differential equation. This is also because you don't have enough tools to solve it in 2U anyway.

In 3U, you develop a useful tool to solving them. If you wish to see the method, you can just post your question again in the 3U thread.
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Solving an equation is not the intended approach here because of the fact they wanted this 'upper bound'. It's not impossible to use it, but it would get incredibly hand-wavy.

[katnisschung]

i need help visualising this...could someone pls draw a diagram thanks

from the top of a vertical cliff the angle of depression to a marker at the base of the cliff is
39 degrees. On the other side of the cliff is another marker, and the angle of depression from
the top of the cliff to this marker is 43 degrees. The markers are 300m apart. Find the height of
the cliff to the nearest metre. (answer 130m)

thanks

[MisterNeo]

i need help visualising this...could someone pls draw a diagram thanks

from the top of a vertical cliff the angle of depression to a marker at the base of the cliff is
39 degrees. On the other side of the cliff is another marker, and the angle of depression from
the top of the cliff to this marker is 43 degrees. The markers are 300m apart. Find the height of
the cliff to the nearest metre. (answer 130m)

thanks

The cliff is like a telegraph pole where it has virtually no thickness and has two faces.

So my way of working is basically:
-Find the two distances from the markers at each side using Sine Rule.
-Find the area of the triangle diagram using that formula on the Ref Sheet.
-Convert it to height using the easy triangle formula, which has a height variable.
Also, you can't just halve the triangles and use Sin Rule because the height is unknown and the base isn't necessarily going to be a perfect 150-150 metres.
Hope this helps

[itssona]

hiii how do i do this
do i take seperate cases

|4x +3| + |x-1| <11

[Natasha.97]

hiii how do i do this
do i take seperate cases

|4x +3| + |x-1| <11

Hi!

Yeah I would do the four cases:
- Pos/Pos
- Pos/Neg
- Neg/Pos
- Neg/Neg

After finding x, test it with the equation to ensure that it holds true (that the value is smaller than 11).

Hope this helps

My solution below:

[RuiAce]

Hi!

Yeah I would do the four cases:
- Pos/Pos
- Pos/Neg
- Neg/Pos
- Neg/Neg

After finding x, test it with the equation to ensure that it holds true (that the value is smaller than 11).

Hope this helps

hiii how do i do this
do i take seperate cases

|4x +3| + |x-1| <11

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[itssona]

ahh thank you Jess and Rui!!

[georgiia]