answer:Let r be the number of red cards and n be the number of black cards in the initial deck. By hypothesis, we have n=2 r. After adding 4 black cards, there are n+4 black cards in the deck, and by hypothesis, we have n+4=3 r. Thus, 4=3 r-n=3 r-2 r=r and n=2 cdot 4=8. Initially, the deck contains r+n=4+8=12 cards.Comment from the graders: The exercise is very well solved! However, be sure to carefully read the instructions: the question here was the total number of cards and not the number of black cards...

answer:For (1), the original expression can be rewritten as (dfrac{3}{2})^{2timesfrac{1}{2}}-1-(dfrac{3}{2})^{-3times(-frac{2}{3})}+dfrac{4}{9}=dfrac{3}{2}-1-dfrac{4}{9}+dfrac{4}{9}=boxed{dfrac{1}{2}}． For (2), the original expression can be rewritten as dfrac{1}{2}log_{3}3+lg (25times4)+2=dfrac{1}{2}+2+2=boxed{dfrac{9}{2}}．

answer:fbox{B}

answer:To find the solution to the given equation, we will first apply the property that if ln(a) = ln(b), then a = b. Hence, we equate the arguments of the natural logarithms:2x + 1 = x^2 - 2.Now, let's arrange the terms in the form of a quadratic equation:x^2 - 2x - 3 = 0.This is a simple quadratic equation which can be factored as follows:(x - 3)(x + 1) = 0.The roots of the equation are x - 3 = 0 and x + 1 = 0. Solving for x, we get:[begin{align*}x - 3 &= 0 Rightarrow x = 3, x + 1 &= 0 Rightarrow x = -1.end{align*}]However, when we substitute x = -1 back into the original logarithmic equation, we get:ln(2(-1) + 1) = ln((-1)^2 - 2),which simplifies toln(-1) = ln(1 - 2),Here we observe that ln(-1) is undefined because the natural logarithm is not defined for negative numbers. Thus, x = -1 is not a valid solution to the original equation.Therefore, the valid solution is:boxed{x = 3}.

answer:4718

answer:1. Claim: The first player wins if and only if ( n = 3 ) or ( n ) is even.2. Base Case: - If ( n = 3 ), the first player wins by splitting the stack into stacks of size ( 1 ) and ( 2 ).3. Definitions: - A *fancy stack* is a stack that can be split (i.e., a stack with at least ( 2 ) pennies). - A *fancy penny* is a penny in any fancy stack. - A *fancy position* is a position such that all of the stacks have an odd number of pennies, and if there is exactly one fancy stack, that stack has more than ( 3 ) pennies.4. Lemma: In any fancy position, the next player (i.e., the player who has his/her turn right now) loses.5. Proof of Lemma: - Assume for contradiction that there is a fancy position such that the next player can win. Let us consider such a position with a minimal number of fancy pennies. - Case 1: This position has at most ( 4 ) fancy pennies. - If there are zero fancy stacks, the previous player has already won. - Otherwise, there can only be one fancy stack, and it must have ( 3 ) pennies, contradicting the assumption that the position is fancy. - Case 2: This position has ( 5 ) fancy pennies. - There can only be one fancy stack, and it must have ( 5 ) pennies. - The next player could split the stack into stacks of size ( 1 ) and ( 4 ), after which the opponent wins by splitting the stack of size ( 4 ) into two stacks of size ( 2 ). - Otherwise, the next player would have to split the stack into stacks of size ( 2 ) and ( 3 ), and then the opponent wins by splitting the stack of size ( 3 ) into stacks of size ( 2 ) and ( 1 ). - Case 3: This position has ( 6 ) fancy pennies. - There must be two fancy stacks, each with ( 3 ) pennies. - The next player must split one of these stacks into stacks of size ( 1 ) and ( 2 ), and then the opponent wins by splitting the other stack of size ( 3 ). - Case 4: This position has at least ( 7 ) fancy pennies. - The next player must split a stack into two stacks, exactly one of which has an even number of pennies. - The opponent should split that stack into a stack of size ( 1 ) and another stack with an odd number of pennies. - All the stacks will have an odd number of pennies again. - The former player creates at most ( 1 ) stack of size ( 1 ) (since the other stack must have an even number of pennies), and the opponent creates either ( 1 ) or ( 2 ) stacks of size ( 1 ). - Thus, the number of fancy pennies decreases by at most ( 3 ), so the new position has at least ( 4 ) fancy pennies remaining, so it must be fancy. - The new position has fewer fancy pennies than the old position, which means that the next player must lose because we assumed that the old position is the minimal fancy position such that the next player can win.6. In either case, we see that the next player must lose. Thus, by contradiction, there exists no fancy position such that the next player can win, as desired.7. Conclusion: - It immediately follows that the first player loses if ( n ) is odd and ( n > 3 ), since the initial position is fancy. - For even ( n ): - If ( n = 4 ), the first player wins immediately by splitting the stack of size ( 4 ) into two stacks of size ( 2 ). - If ( n > 4 ), the first player wins by splitting the stack into stacks of size ( 1 ) and ( n-1 ), creating a fancy position that causes the opponent to lose.(therefore) The first player wins if and only if ( n = 3 ) or ( n ) is even.The final answer is ( boxed{ n = 3 } ) or ( n ) is even.