MaxWong

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0.83 = 5/6

Therefore

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0.083 = 5/6 /10 = 5/60 = 1/12

Therefore

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3.083 = 3 + 5/60 = 3 1/12

I guess you mean \(\dfrac{24p^5}{6p^2}\)

\(\dfrac{24p^5}{6p^2}\\=4p^3\)

Because 24/6 = 4 and p^5/p^2 = p^(5-2) = p^3

And I think the notation for chain functions is \((f\circ g)(x)\)?

Luckily I am still correct in question 2, I just multiply out (x+6)^2 and divide all the terms by 2 to get rid of the fraction thing and made it ax^2 + bx + c.

:)

people who are needing help with math also can post a new question :)

2300.0 probably? :)

Still my unreliable knowledge, sorry for that again.

Part A:

Force = mass x acceleration,

Force is doubled, mass constant, acceleration doubles.

a = 2,4 x 2 = 4,8 m/s2

Part B:

Still the same formula.

Force is constant, mass doubled, acceleration halves.

a = 2.4/2 = 1.2 m/s2

often_wrong you are probably correct this time :), we got the same answer, we will be not so likely that we are wrong together :)

Still my shaky knowledges, I apologize again for this.

Part A:

Force is halved, Mass remains unchanged, Acceleration is halved.

So that a = 5 m/s2

Part B:

Force is constant, mass is halved, acceleration is doubled.

Therefore a = 20 m/s2

Part C;

Force is halved, Mass is halved, therefore acceleration is constant.

Therefore a = 10 m/s2

Part D:

Force is halved, Mass is doubled, Acceleration is halved twice <---- which is 1/4.

Therefore a = 5 x 1/4 = 1 1/4 m/s^2

I have just started to learn Physics just a month ago and my concepts may be wrong. I apologize for this first.

Part A.

B because force = mass x acceleration so the mass is bigger than others so the accelaration should be smaller.

Part B.

C, same reason.

Part C.

Because the force remains the same, we have:

\(\text{Force} = 5 \times \text{Mass}_{\text{A}}\\ \text{and}\\ \text{Force} = 3 \times \text{Mass}_{\text{B}}\)

So that we have:
\(5\times \text{Mass}_{\text{A}} = 3 \times \text{Mass}_{\text{B}}\\ \dfrac{\text{Mass}_{\text{A}}}{\text{Mass}_{\text{B}}}=\dfrac{3}{5}\\ \text{Mass}_{\text{A}} : \text{Mass}_{\text{B}}= 3:5\)

I have got that question by my one-month learning knowledge :)

I just came up with a proof. Apologies to everyone.

\(\ln(\sqrt{-1})=\dfrac{\ln(-1)}{2}\\ e^{i\pi}+1=0\\ e^{i\pi}=-1\\ i\pi = \ln(-1)\\ \therefore \ln(\sqrt{-1})=\dfrac{\ln(-1)}{2}=\dfrac{i\pi}{2}\)