answer：First, I need to determine how many bones Lassie had before eating half of them on Saturday. She ended up with 35 bones after receiving 10 more on Sunday. This means that before receiving the 10 bones, she had 35 - 10 = 25 bones.Since she ate half of her bones on Saturday, the number of bones she had before eating them would be twice the amount she had on Sunday morning. Therefore, multiplying the number of bones she had on Sunday by 2 gives the original number of bones: 25 * 2 = 50.Thus, Lassie started with 50 bones before eating them on Saturday.

answer：First, I need to determine the total number of ways to draw 4 balls from the 15 balls in order. Since each draw reduces the number of available balls by one, the total number of unrestricted draws is 15 × 14 × 13 × 12.Next, to satisfy the condition that at least one of the drawn balls is even-numbered, I'll use the complementary counting principle. This means I'll first calculate the number of ways to draw four odd-numbered balls. There are 8 odd-numbered balls (1, 3, 5, 7, 9, 11, 13, 15), so the number of ways to draw four odd balls is 8 × 7 × 6 × 5.Finally, the number of valid drawings that include at least one even-numbered ball is the total number of unrestricted draws minus the number of all-odd draws. This gives me the final count of valid ways.

answer：Alright, so I've got this problem here about an athlete jumping into the water. Let me try to understand what's going on. The athlete jumps from a height of 10 meters. At some point during the jump, when they're at 5 meters above the water, they're moving at a speed of ( v ). Then, when they hit the water, their speed is ( 2v ). Hmm, that seems interesting. I wonder if that's possible.Okay, first, I need to recall some physics principles that might help here. I think this has to do with motion under gravity and maybe energy conservation. When you jump from a height, you accelerate due to gravity, and your speed increases as you fall. So, the higher you are, the more potential energy you have, which converts into kinetic energy as you fall.Let me write down what I know:- Initial height (( h_1 )) = 10 meters- Height at which speed is ( v ) (( h_2 )) = 5 meters- Speed at ( h_2 ) = ( v )- Speed at water level (( h_3 )) = ( 2v )I need to find out if this scenario is possible. Maybe I can use the equations of motion or energy conservation to check.Starting with energy conservation. The total mechanical energy (potential + kinetic) should remain constant if we ignore air resistance. So, at the highest point (10 meters), all the energy is potential. As the athlete falls, this potential energy converts into kinetic energy.Let's denote:- ( m ) = mass of the athlete (it will cancel out, so maybe I don't need to worry about it)- ( g ) = acceleration due to gravity (( 9.8 , text{m/s}^2 ))- ( v_1 ) = initial velocity at 10 meters (which is 0, since they start from rest)- ( v_2 ) = velocity at 5 meters (( v ))- ( v_3 ) = velocity at water level (( 2v ))Using the conservation of energy between the initial point (10 meters) and the point at 5 meters:[ mgh_1 = frac{1}{2}mv_2^2 + mgh_2 ]Simplifying, since mass ( m ) cancels out:[ gh_1 = frac{1}{2}v_2^2 + gh_2 ]Plugging in the values:[ g times 10 = frac{1}{2}v^2 + g times 5 ][ 10g = frac{1}{2}v^2 + 5g ]Subtracting ( 5g ) from both sides:[ 5g = frac{1}{2}v^2 ]Multiplying both sides by 2:[ 10g = v^2 ]So,[ v = sqrt{10g} ]Okay, so the velocity at 5 meters is ( sqrt{10g} ).Now, let's apply energy conservation from 5 meters to the water level:[ frac{1}{2}mv_2^2 + mgh_2 = frac{1}{2}mv_3^2 ]Again, mass ( m ) cancels out:[ frac{1}{2}v^2 + gh_2 = frac{1}{2}v_3^2 ]We know ( v_3 = 2v ), so:[ frac{1}{2}v^2 + g times 5 = frac{1}{2}(2v)^2 ]Simplifying the right side:[ frac{1}{2}v^2 + 5g = frac{1}{2} times 4v^2 ][ frac{1}{2}v^2 + 5g = 2v^2 ]Subtracting ( frac{1}{2}v^2 ) from both sides:[ 5g = frac{3}{2}v^2 ]Multiplying both sides by ( frac{2}{3} ):[ frac{10}{3}g = v^2 ]So,[ v = sqrt{frac{10}{3}g} ]Wait a minute, earlier I found ( v = sqrt{10g} ), and now I'm getting ( v = sqrt{frac{10}{3}g} ). That doesn't make sense. There's a contradiction here. How can ( v ) have two different values?Maybe I made a mistake in my calculations. Let me double-check.Starting from the first part:[ gh_1 = frac{1}{2}v^2 + gh_2 ][ 10g = frac{1}{2}v^2 + 5g ]Subtracting ( 5g ):[ 5g = frac{1}{2}v^2 ][ v^2 = 10g ][ v = sqrt{10g} ]That seems correct.Now, the second part:[ frac{1}{2}v^2 + 5g = frac{1}{2}(2v)^2 ][ frac{1}{2}v^2 + 5g = 2v^2 ]Subtracting ( frac{1}{2}v^2 ):[ 5g = frac{3}{2}v^2 ][ v^2 = frac{10}{3}g ][ v = sqrt{frac{10}{3}g} ]Hmm, so according to the first calculation, ( v = sqrt{10g} ), and according to the second, ( v = sqrt{frac{10}{3}g} ). These two results should be the same if the scenario is possible, but they're not. That suggests that the assumption that the athlete enters the water with a speed of ( 2v ) is inconsistent with the conservation of energy.Maybe I should think about this differently. Perhaps there's another factor at play, like air resistance? But the problem doesn't mention air resistance, so I think we're supposed to ignore it.Alternatively, maybe the athlete isn't just in free fall. Perhaps they're actively moving, like jumping or pushing off the board, which could add energy to the system. But the problem states they jump into the water from a height of 10 meters, which sounds like they're just falling, not adding any extra energy.Wait, another thought: if the athlete is entering the water with a speed of ( 2v ), that's double the speed at 5 meters. But according to the energy calculations, the speed should increase as they fall, but not necessarily double. Maybe the relationship between the heights and speeds isn't linear, so doubling the speed might not align with the energy conservation.Let me recall the kinematic equations. The velocity squared is proportional to the distance fallen when starting from rest. The formula is:[ v^2 = u^2 + 2as ]Where ( u ) is the initial velocity, ( a ) is acceleration, and ( s ) is the distance.In this case, starting from rest, ( u = 0 ), so:[ v^2 = 2as ]Here, ( a = g ), and ( s ) is the distance fallen.From 10 meters to 5 meters, the distance fallen is 5 meters. So:[ v^2 = 2g times 5 = 10g ]Which matches my earlier result, ( v = sqrt{10g} ).From 5 meters to the water, another 5 meters fallen:[ v_{text{water}}^2 = v^2 + 2g times 5 = 10g + 10g = 20g ]So,[ v_{text{water}} = sqrt{20g} ]But according to the problem, ( v_{text{water}} = 2v ). Let's see what ( 2v ) is:[ 2v = 2sqrt{10g} ][ (2v)^2 = 4 times 10g = 40g ]But from the kinematic equation, ( v_{text{water}}^2 = 20g ). So,[ 20g neq 40g ]That's not equal. Therefore, the speed at water entry should be ( sqrt{20g} ), not ( 2sqrt{10g} ). Hence, the scenario described in the problem is not possible under the assumption of free fall and conservation of energy.Maybe the athlete is doing something else, like pushing off the water or something, but the problem doesn't mention that. So, based on the given information and assuming free fall, it's not possible for the athlete to enter the water at ( 2v ) if they passed 5 meters at speed ( v ).I think that's the conclusion. The problem's conditions contradict the laws of physics as we understand them in this context.

answer：Okay, so I have this problem here about A and B's money. Let me try to understand what it's asking. First, it says, "The sum of six times A's money and B's money equals 66.00." So, if I let a represent A's money and b represent B's money, then this translates to the equation:6a + b = 66Got that. So, that's my first equation.Then, the second part says, "Four times A's money plus B's money is less than 48.00." Okay, so that would be:4a + b < 48Alright, so now I have two equations:1. 6a + b = 662. 4a + b < 48I need to solve for a and b. Hmm, let's see. Maybe I can solve the first equation for one variable and substitute into the second inequality.Let me solve the first equation for b. So, from 6a + b = 66, subtracting 6a from both sides gives:b = 66 - 6aOkay, so now I can substitute this expression for b into the second inequality. Let's do that.So, replacing b in 4a + b < 48 with (66 - 6a):4a + (66 - 6a) < 48Now, let's simplify this. Combining like terms:4a - 6a + 66 < 48That simplifies to:-2a + 66 < 48Now, I need to solve for a. Let's subtract 66 from both sides:-2a < 48 - 66Which is:-2a < -18Now, to solve for a, I'll divide both sides by -2. But wait, I remember that when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips. So, dividing both sides by -2:a > (-18)/(-2)Which simplifies to:a > 9Okay, so a is greater than 9. That gives me a lower bound for a. Now, what about b?Well, from the first equation, we had:b = 66 - 6aSince a > 9, let's see what happens to b. If a is just slightly more than 9, say 9.1, then:b = 66 - 6*(9.1) = 66 - 54.6 = 11.4But as a increases, b decreases. For example, if a is 10:b = 66 - 6*10 = 66 - 60 = 6If a is 11:b = 66 - 6*11 = 66 - 66 = 0Wait, b can't be negative, right? So, b has to be at least 0. That means a can't be more than 11 because at a=11, b=0. So, a is between 9 and 11, and b is between 0 and 6.But looking back at the options, let's see:A) a > 10, b = 6B) a > 9, b < 6C) a > 9, b = 6D) a > 9, b > 6E) 2a = 3bHmm. From my calculations, a > 9 and b = 6 when a=10, but b can be less than 6 as a increases beyond 10. So, when a is exactly 10, b is exactly 6, but for a >10, b becomes less than 6. So, the most accurate statement would be a >9 and b <6 when a>10, but when a=10, b=6.Looking at the options, option C says a >9, b=6, which is only true when a=10. But since a can be greater than 9, not necessarily exactly 10, and b can be less than 6, option B says a >9, b <6, which is also partially true but doesn't account for the case when a=10, b=6.Option A is a >10, b=6, which is only true if a is exactly 10, but a can be greater than 9, not necessarily greater than 10. So, that's too restrictive.Option D says a >9, b >6, which contradicts our earlier finding that b decreases as a increases beyond 9, so b would be less than 6 when a is greater than 10.Option E is 2a=3b. Let's check if that's true. If a=10, b=6, then 2*10=20 and 3*6=18, which are not equal. So, that's not correct.So, the best option is C) a >9, b=6, but only when a=10. However, since a can be greater than 9, and b can be less than 6, I think the most accurate answer is B) a >9, b <6, but considering that when a=10, b=6, which is a specific case.Wait, but the options don't have a combination of a >9 and b ≤6. So, perhaps the intended answer is C) a >9, b=6, considering that when a=10, b=6, and that's a specific solution, but a can be greater than 9, making b less than 6.I think the answer is C) a >9, b=6, but I'm a bit confused because b can be less than 6 when a is greater than 10. Maybe the question expects the relationship when a is exactly 10, making b=6, so the answer is C.

answer：First, I need to find the value of ( y ) when ( x = -7 ). I'll substitute ( x = -7 ) into the equation ( x^2 = y - 5 ).[(-7)^2 = y - 5 implies 49 = y - 5]Next, I'll solve for ( y ) by adding 5 to both sides of the equation.[y = 49 + 5 = 54]Now, I'll find the value of ( y ) when ( x = 2 ). I'll substitute ( x = 2 ) into the equation ( x^2 = y - 5 ).[2^2 = y - 5 implies 4 = y - 5]Finally, I'll solve for ( y ) by adding 5 to both sides of the equation.[y = 4 + 5 = 9]

answer：Alright, so I've got these two algebraic expressions to simplify, and I'm a bit nervous because I'm still getting the hang of this. Let's take it step by step.Starting with the first one: 2x - 3(x - 1). Okay, so I need to simplify this. I remember that simplifying usually involves distributing and combining like terms. Let me try to recall how distribution works. If I have something like a(b + c), it becomes ab + ac. So, applying that here, the -3 needs to be distributed to both x and -1 inside the parentheses.Wait, hold on. The expression is 2x minus 3 times (x minus 1). So, it's 2x - 3(x - 1). To distribute the -3, I need to multiply both x and -1 by -3. So, -3 times x is -3x, and -3 times -1 is +3. So, after distributing, the expression becomes 2x - 3x + 3.Now, I need to combine like terms. The terms with x are 2x and -3x. Combining those, 2x minus 3x is -x. So, the expression simplifies to -x + 3. Hmm, usually, we write expressions with the variable term first, so maybe it's better to write it as 3 - x. Yeah, that seems more standard.Okay, that wasn't too bad. Now, moving on to the second expression: 6(ab² - a²b) - 2(3a²b + ab²). This looks a bit more complicated because there are multiple variables and exponents involved. Let's see if I can handle this.First, I need to distribute the 6 and the -2 across their respective parentheses. Starting with the first part: 6 times (ab² - a²b). Distributing the 6, I get 6ab² - 6a²b. Okay, that part makes sense.Now, the second part is -2 times (3a²b + ab²). Distributing the -2, I get -6a²b - 2ab². So, putting it all together, the expression becomes 6ab² - 6a²b - 6a²b - 2ab².Next, I need to combine like terms. Let's look at the ab² terms first. I have 6ab² and -2ab². Combining those, 6ab² minus 2ab² is 4ab². Now, for the a²b terms, I have -6a²b and -6a²b. Combining those, -6a²b minus 6a²b is -12a²b.So, after combining like terms, the expression simplifies to 4ab² - 12a²b. That seems to be as simplified as it can get. I don't think there are any further like terms to combine or any other distributions needed.Wait a second, let me double-check my work. Did I distribute the -2 correctly? Yes, -2 times 3a²b is -6a²b, and -2 times ab² is -2ab². Okay, that's correct. And combining the terms: 6ab² - 2ab² is indeed 4ab², and -6a²b - 6a²b is -12a²b. Looks good.I think I've got both expressions simplified correctly. The first one is 3 - x, and the second one is 4ab² - 12a²b. I feel more confident about the first one, but the second one seems right too. Maybe I can plug in some numbers to verify.For the first expression, let's say x = 2. Original expression: 2(2) - 3(2 - 1) = 4 - 3(1) = 4 - 3 = 1. Simplified expression: 3 - 2 = 1. They match. Good.For the second expression, let's choose a = 1 and b = 2. Original expression: 6(1*2² - 1²*2) - 2(3*1²*2 + 1*2²) = 6(4 - 2) - 2(6 + 4) = 6(2) - 2(10) = 12 - 20 = -8.Simplified expression: 4*1*2² - 12*1²*2 = 4*4 - 12*2 = 16 - 24 = -8. They also match. Okay, that gives me more confidence that my simplifications are correct.I think I've got it!