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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.

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

1.\(\quad (f\cdot g)(x)\\=(x^2-7x+12)(\dfrac{7}{x^2}-16)\\ = 7 - \dfrac{49}{x}+\dfrac{84}{x^2}-16x^2+112x-192\\ =-16x^2+112x-185-\dfrac{49}{x}+\dfrac{84}{x^2}\)

2.\(f(x)=\sqrt{2x}-6\\ y=\sqrt{2x}-6\\ x = \sqrt{2y}-6\\ \sqrt{2y}=x+6\\ 2y = x^2 + 12x + 36\\ y = \dfrac{x^2}{2}+6x+18\\ f^{-1}(x)=\dfrac{x^2}{2}+6x+18\)

How about this :/

\(let\\ \begin{align} x&=ln\; i\\ e^x&=i\\now\\ i&=cos\frac{\pi}{2}+isin\frac{\pi}{2}\\\therefore\\ i&=e^{(\pi/2)i}\\\therefore \\ e^x&=e^{(\pi/2)i}\\\therefore\\ x&=\frac{\pi }{2}\;i \\\therefore\\ ln\;i&=\frac{\sqrt{-\pi^2}}{2} \end{align} \)

Really? It doesn't work for me!

Anyone???

Good and you but math only pleaese

\(\begin{array}{rll}4x^2+15&=&x^2+18x\\3x^2-18x+15&=&0\\x^2-6x+5&=&0\\(x-1)(x-5)&=&0\end{array}\\\text{Either one of the factors equal 0.}\\\begin{array}{rl}x-1=0&x-5=0\\x=1&x=5\\\end{array}\)

\(\dfrac{a}{b}\times c = \dfrac{a\times c }{b}\).

Examples: 

\(\dfrac{5}{7} \times 2 = \dfrac{5\times 2}{7}= \dfrac{10}{7}=1\dfrac{3}{7}\)

\(\dfrac{2}{3}\times 15 = \dfrac{30}{3}=10\)

\(\dfrac{4}{9}\times 32 = \dfrac{128}{9} = 14\dfrac{2}{9}\)

Hi Alan,

I have been trying to work out what you have said.

I understand your example of  course but I do not understand why that means that this is the only answer.  How can you take just one example and say the answer is always going to be the same.??

Let r(x)=F(g(h))),      where h(1)=2, g(2)=5, h'(1)=4, g'(2)=4, and f'(5)=6. Find r'(1).

I think this is right.   

\(\begin{align} \frac{dr}{dx}&=\frac{df}{dg} \cdot \frac{dg}{dh}\cdot \frac{dh}{dx}\\ \end{align} \)

 I'm told that  when x=1     dh/dx =2   but I don't know from the info given what the other two derivatives are.  How can I use all that other information without resorting to a specific example ???

22521008071930811247210364724710555878336746262578603045969925706618268298469415658031134405487.2423412144568778 according to the calculator. Just copy and paste.

A. a=4.8m/s/s

B. a=1.2m/s/s

4x^2+15=x^2+18x

• minus 4x^2 from both sides

15=-3x^2+18x

• minus 15 from both sides

-3x^2+18x-15=0

• take a factor of -3 out of brackets

-3(x^2-6x+5)=0

• sum and product gives factors of 6 and -1
• factorise the brackets

-3(x+6)(x-1)=0

• Null factor theorem means:

x = -6 or 1

what is 1% of 1.6 billion

1/100 x 1.6 x 10^9 =16,000,000 =1.6 x 10^6

2cm*2cm*2cm = 8cm³

8cm*12cm*10cm = 960cm³

960:8 = 120

120>100

The answer is yes.

What is x+30 Drgress? Cause I Need to know please so i dont get it wrong

"x + 30 Drgress" is to say "x + 30 degrees" or "x + 30 ° "

!

Would 7 to the 3rd power over 7 to the 4th power equal 16807?

\({\color{blue}7^3\times 7^2 = 7^5=16807}\)  !

\({\color{blue}\frac {7^3}{7^4}} {\color{red}\times \ 7^6} {\color{blue}= 16807}\)

21 brah

21 brah

Like this :))

(7^3)/(7^4)  →  7^(3-4)  →  7^(-1)  → 1/7

. 

1/2^-3  →  2^3 →  8

.

Rate of change between (x1, y1) and (x2, y2) is  (y2 - y1)/(x2 - x1)

.

See the answer given here: http://web2.0calc.com/questions/need-help_79616#r4 

.