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That means {\(\dfrac{a}{1-r} = 2\\ \dfrac{a^2}{1-r^2} = 3\).

\(\dfrac{2a}{1+r} = 3\\ \dfrac{2(2-2r)}{1+r} = 3\\ 4-4r = 3+3r\\ 7r = 1\\ r = \dfrac{1}{7}\\ \dfrac{2a}{\frac{8}{7}} = 3\\ 7a = 12\\ a =\dfrac{12}{7}\\\)

So \((r,a) = \left(\dfrac{1}{7},\dfrac{12}{7}\right)\)

10% of 60% = 0.1 * 0.6 = 0.06

20% of 60% = 2(10% of 60%) = 0.12

25% of 60% = 0.06 * 5 / 2 = 0.15

50% of 60% = 2*(25% of 60%) = 0.3

https://web2.0calc.com/questions/given-that-f-x-x-k-where-k-gt-0-what-is-the-range#r2

\(\begin{align}&306\\&---\\15&|4598\\&\;45\\&---\\&\quad\;98\\&\quad\;90\\&---\\&\quad\quad\! 8\end{align}\)

What a terrible question.

\(\bar{a} = \dfrac{|v| - |v_0|}{\Delta t} = \dfrac{1000-0~m/s}{0.001s} = 10^6 ~m/s^2\)

Ok, that's straight forward enough but the reason that the question is terrible is that 

it suggests that you can actually accelerate this bullet up to the speed of light.

I realize it's not a course on relativity but there's no reason to suggest that the impossible can be done.

The real answer to this question is that it will take forever the bullet to reach the speed of light because 

it cannot be reached by any sort of matter.

I suppose the answer they are looking for is

\(t = \dfrac{300 \times 10^6~m/s}{10^~m/s^2} = 300~s = 5~min\)

The X intercepts of the parabola,

y= 2x^2 + 13x - 7

are at (p,0) and (q,0) Find P and Q

You need to give us more information - an example to work with.

Try graphing both you graphs here.

Thanks Heureka, I did not know how to do that :)

Yes, it seems all polygons are circles in disguise :)

What a great discovery you have made   

(a,b)=?

1. a(-5+4i)+b(-5-4i)=38

2. a(7+2i)+b(7-2i)=38

=> a=1-6i and b=1+6i