Rom

3 2D vectors cannot be linearly independent.

By eye

a=2

b=-3

c=7

5/7

deleted

2 = 2+0

3 = 3+0

4 = 3+1

5 = 5+0

6 = 2+4, 5+1

7 = 3+4

8 = 7+1

9 = 7+2

10 = 

2+8 X

3+7 X

5+5 X

7+3 X

10 is the smallest number > 1 that can't be expressed as stated.

\(P = P[Q \cup J | K]P[K]+ P[J|Q]P[Q]\\ P=\dfrac{8}{11}\dfrac 1 3 + \dfrac{4}{11}\dfrac 1 3 = \dfrac{12}{33} = \dfrac{4}{11}\)

Another way we can do this is to note that the number of pairs where card 2 > card 1

is equal to the number of pairs where card 1 > card 2 and that these two plus the number

of pairs where cards 1 and 2 are equal equal the total number of pairs.

\(P[\text{card 1 = card 2}] = 3 \cdot \dfrac 1 3 \cdot \dfrac{3}{11} = \dfrac{3}{11}\\ 1 = \dfrac{3}{11} + 2 P\\ 2P = \dfrac{8}{11}\\ P=\dfrac{4}{11}\)

this proves nothing

\(\text{We start with the acceleration to find the velocity}\\ v(t) = \displaystyle \int_0^t a(\tau) d\tau + v_0 = t^2+3t-9\\~\\ s(t) = \displaystyle \int_0^t v(\tau)d\tau + s_0 = \dfrac{t^3}{3}+\dfrac 3 2 t^2 - 9t + 6\)

If you mean by "have the same graph" that you can pick up the graph of one polynomial and place it exactly on top of the other one, 

i.e. that p1(x) = p2(x) then yes they have the same roots.

As the other poster noted the converse is not true.  Polynomials with identical roots can have different graphs.

set the Gradient equal to zero and solved (with the help of software) and examined the solutions to see which 

produced a maximum (vice a minimum) and what it was.

There's probably some more clever geometric solution here.

\(\text{The answer appears to be $\dfrac 1 2(1+\sqrt{5})$}\)

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