Mathematics Question Thread

[katnisschung]

i thought using the product rule was correct but apparently not.
q and a attached

[RuiAce]

i thought using the product rule was correct but apparently not.
q and a attached

[jamonwindeyer]

i thought using the product rule was correct but apparently not.
q and a attached

It would probably be slightly easier to expand that inside bit first, just to show another way

[katnisschung]

thanks jamon yeah i did expand it haha i couldn't wrap my head around simplifying it 
i have another question regarding differentiating logs (we had a sub for this and i didn't understand anything)

i tried differentiating the attached by applying change of base rule
and then did quotient rule (yeah i think i'm overthinking it)
can someone provide an explanation as to how to differentiate logs with a base other than e (thats all i remember being taught)

[RuiAce]

thanks jamon yeah i did expand it haha i couldn't wrap my head around simplifying it 
i have another question regarding differentiating logs (we had a sub for this and i didn't understand anything)

i tried differentiating the attached by applying change of base rule
and then did quotient rule (yeah i think i'm overthinking it)
can someone provide an explanation as to how to differentiate logs with a base other than e (thats all i remember being taught)

[laurenf58]

How do I find the focal length of the parabola x^2=8y-24? Thanks!

[jakesilove]

How do I find the focal length of the parabola x^2=8y-24? Thanks!

Hey! Recall that the focal length, 'a', is defined as when the equation of the parabola is



x and y can be shifted. In this case,



Clearly,



So a=2

[mcheema]

Can someone pls help me with q26

[anotherworld2b]

Can I please have some help with these 2 questions please? I got stuck on 20 d) and the questions onwards

[kiiaaa]

hi all
i have a question and im soo confused how to do it especially when i look at how my teacher did im like wot?!?. could you please explain me the process in how to do and approach the question
thanks
QUESTION: find the equation to the tangent to the curve y'= 1+2e^2x
MY TEACHERS SOLUTION: in the attactment below (i hope you could understand that dodgy working)
thanks once again

[RuiAce]

hi all
i have a question and im soo confused how to do it especially when i look at how my teacher did im like wot?!?. could you please explain me the process in how to do and approach the question
thanks
QUESTION: find the equation to the tangent to the curve y'= 1+2e^2x
MY TEACHERS SOLUTION: in the attactment below (i hope you could understand that dodgy working)
thanks once again

There is no attachment.

Also was it actually given y'=1+2e^2x and not y=1+2e^2x?

[jamonwindeyer]

Can I please have some help with these 2 questions please? I got stuck on 20 d) and the questions onwards

Sure!

For 20E, you know the total is six. With 16 possible outcomes (some identical), the ones that match this are 24, 42, 33 and 33.  Only one of these has 4 as the second spin. Meaning the answer is just the probability of spinning a 2 and then a 4 (use whatever value of \(k\) you obtained).

For 20F, product rule of probability - Multiply the probability of a 4, with the probability of  3, with the probability of a 2.

For 20G, consider the possible ways we can arrange the outcomes from 20F - There are \(3!=6\) arrangements. So we multiply the answer from 20F by 6!

20H, approach it in the same way as 20E.

20I, calculate the probabilites of three 1's, three 2's, three 3's and three 4's and add it together

For 21, do it piece by piece. First, what are the odds of choosing no faulty components? Well it means we pick four healthy components in a row:



For 1 faulty, we do a similar thing, multiplying by the number of ways we could possibly arrange the choices:



Do similar maths to find \(P_2,P_3\) and \(P_4\) - They will form your probability distribution (remember 5dp as specified!)

[jamonwindeyer]

(Image removed from quote.)

Can someone pls help me with q26

Hey! I've attached my working below - Pretty sure there is a more efficient way to find \(BF\), but this method only uses pythagoras and basic ideas of similarity

Working

[anotherworld2b]

for 20E would it be 0.4 x 0.3 = 0.12?
but the answer says it is 0.48
for 20H would it be
442, 424, 244, 334, 343, 433, = 6 possible outcomes
i'm not quite sure what to do from here.
I tried q21 but when i multiplied the values for P1 i got 1.848 but the answers says 0.30808

Sure!

For 20E, you know the total is six. With 16 possible outcomes (some identical), the ones that match this are 24, 42, 33 and 33.  Only one of these has 4 as the second spin. Meaning the answer is just the probability of spinning a 2 and then a 4 (use whatever value of \(k\) you obtained).

For 20F, product rule of probability - Multiply the probability of a 4, with the probability of  3, with the probability of a 2.

For 20G, consider the possible ways we can arrange the outcomes from 20F - There are \(3!=6\) arrangements. So we multiply the answer from 20F by 6!

20H, approach it in the same way as 20E.

20I, calculate the probabilites of three 1's, three 2's, three 3's and three 4's and add it together

For 21, do it piece by piece. First, what are the odds of choosing no faulty components? Well it means we pick four healthy components in a row:



For 1 faulty, we do a similar thing, multiplying by the number of ways we could possibly arrange the choices:



Do similar maths to find \(P_2,P_3\) and \(P_4\) - They will form your probability distribution (remember 5dp as specified!)

[jamonwindeyer]

for 20E would it be 0.4 x 0.3 = 0.12?
but the answer says it is 0.48
for 20H would it be
442, 424, 244, 334, 343, 433, = 6 possible outcomes
i'm not quite sure what to do from here.
I tried q21 but when i multiplied the values for P1 i got 1.848 but the answers says 0.30808

Not quite, you need to adjust for the sample space because you KNOW the sum is 4, so you are actually only taking one quarter of your sample space (4 outcomes out of 16), so multiply your answer by 4!

For 20H, calculate the probability of each and add all of them together.

Woops, for P1 it is just 4, not 4 factorial (You can pick the faulty unit first, second, third or fourth - Four possible arrangements) that should hopefully fix it? Don't have my calculator on me so can't check