Answer:
Hope this helps :) If it doesn't I can see what I can do
Since the exponent is on the outside of the paranthesis, you'd have to multiply it by everything on the inside: a^m x b^m
Using Laplace transform we have:L(x')+7L(x) = 5L(cos(2t))sL(x)-x(0) + 7L(x) = 5s/(s^2+4)(s+7)L(x)- 4 = 5s/(s^2+4)(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)
=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)
now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:(7-s)/[53(s^2+4)] + 1/53(s+7). So:
L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
denoting T:= L^(-1)and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)
Answer:
47 rounds and Eve wins.
Step-by-step explanation:
If read right, the highest player, (player with the most tokens) has to lose 3 tokens each round, so starting off with Evelyn, 17 - 3 = 14, and the other two each have +1 added to them, so, E=14, A=17 V=16, now subtract three from A (Amelia) and don't forget that one token is always being discarded each round. So the toal number of tokens at the start was 48, (17 + 16 + 15) so right now the total tokens should be 47, which it is. now starting the second round, A gives up 3 tokens so, 17 -3, again equals 14 and each of the two other players get one token, so E=15, A=14 V=17. Now you know the general trend, so we can figure this out seeing that after each round the number of tokens decrease by one, so there will be a round till all tokens are out except one, so there will be 47 rounds and since E started with the highest number they will win.
Answer:
2x+3
Step-by-step explanation:
Let x be the number
twice a number
2x
3 more than
2x+3
