answer：Alright, so I have this equation here: (m - 3)x^{m² - 2m - 1} - mx + 6 = 0. It says it's a quadratic equation, and I need to find the coefficient of the linear term. The options are A: -1, B: 1, C: 3, D: 3 or -1.First, let me recall what defines a quadratic equation. A quadratic equation is a polynomial of degree 2, which means the highest power of x should be 2. So, in this equation, the term with x raised to some exponent is (m - 3)x^{m² - 2m - 1}. For this to be a quadratic equation, the exponent m² - 2m - 1 must equal 2 because x² is the highest degree term.So, I need to set up the equation m² - 2m - 1 = 2 and solve for m. Let's do that step by step.Starting with:m² - 2m - 1 = 2Subtract 2 from both sides to bring all terms to one side:m² - 2m - 1 - 2 = 0Simplify:m² - 2m - 3 = 0Now, I have a quadratic equation in terms of m: m² - 2m - 3 = 0. I can try to factor this. Let me see if I can find two numbers that multiply to -3 and add up to -2.Hmm, the factors of -3 are (1, -3) and (-1, 3). Let's test these:1 * (-3) = -3 and 1 + (-3) = -2. Perfect!So, the equation factors as:(m - 3)(m + 1) = 0Setting each factor equal to zero gives the solutions:m - 3 = 0 => m = 3m + 1 = 0 => m = -1So, m can be either 3 or -1. But wait, I need to make sure that the coefficient of x² isn't zero because if m - 3 = 0, then the x² term would disappear, and the equation wouldn't be quadratic anymore. So, let's check m = 3.If m = 3, then the coefficient of x² is (3 - 3) = 0. That means the equation becomes 0x² - 3x + 6 = 0, which simplifies to -3x + 6 = 0. That's a linear equation, not quadratic. So, m = 3 is not a valid solution because it doesn't satisfy the condition of the equation being quadratic.Therefore, the only valid solution is m = -1.Now, let's substitute m = -1 back into the original equation to find the coefficient of the linear term.Original equation:(m - 3)x^{m² - 2m - 1} - mx + 6 = 0Substitute m = -1:(-1 - 3)x^{(-1)² - 2*(-1) - 1} - (-1)x + 6 = 0Simplify each part step by step.First, calculate the coefficient of x²:(-1 - 3) = -4Next, calculate the exponent:(-1)² = 1-2*(-1) = 2So, exponent is 1 + 2 - 1 = 2So, the x² term is -4x².Now, the linear term is -mx. Since m = -1, this becomes -(-1)x = x. So, the coefficient of x is 1.The constant term is 6.Putting it all together, the equation becomes:-4x² + x + 6 = 0So, the coefficient of the linear term is 1.Let me double-check my steps to make sure I didn't make a mistake.1. Set the exponent equal to 2: m² - 2m - 1 = 22. Solved the quadratic equation: m² - 2m - 3 = 03. Factored it into (m - 3)(m + 1) = 0, giving m = 3 or m = -14. Discarded m = 3 because it makes the coefficient of x² zero, which isn't allowed for a quadratic equation5. Substituted m = -1 into the original equation6. Calculated each term and found the coefficient of x to be 1Everything seems to check out. So, the coefficient of the linear term is 1, which corresponds to option B.**Final Answer**boxed{B}

answer：Okay, so I need to simplify the expression ( t^3 cdot t^4 ). Hmm, I remember that when you multiply variables with exponents, there's a rule about adding the exponents. Let me think about that. First, I know that ( t^3 ) means ( t ) multiplied by itself three times, right? So, ( t^3 = t cdot t cdot t ). Similarly, ( t^4 ) is ( t cdot t cdot t cdot t ). So, if I multiply ( t^3 ) and ( t^4 ) together, I'm essentially multiplying seven ( t )'s together because 3 plus 4 is 7. Wait, is that always the case? Like, if I have ( t^a cdot t^b ), does it always equal ( t^{a+b} )? I think so. Yeah, that sounds familiar from what I've learned before. So, applying that rule here, ( t^3 cdot t^4 ) should be ( t^{3+4} ), which is ( t^7 ). Let me double-check that to make sure I'm not making a mistake. If I expand both expressions, ( t^3 ) is ( t cdot t cdot t ) and ( t^4 ) is ( t cdot t cdot t cdot t ). Multiplying them together gives me ( t cdot t cdot t cdot t cdot t cdot t cdot t ), which is seven ( t )'s multiplied together. So, that's ( t^7 ). I guess another way to think about it is using the properties of exponents. When you multiply like bases, you add the exponents. So, ( t^3 cdot t^4 = t^{3+4} = t^7 ). Yeah, that makes sense. I wonder if there are any exceptions to this rule. Like, does it only work when the base is the same? For example, if I had ( t^3 cdot s^4 ), could I still add the exponents? No, I don't think so because the bases are different. In that case, I can't combine them like that. So, it's important that both terms have the same base for this rule to apply. Also, what if one of the exponents is zero? I remember that anything raised to the power of zero is 1. So, if I had ( t^3 cdot t^0 ), that would be ( t^{3+0} = t^3 ), which is correct because ( t^0 = 1 ), and multiplying by 1 doesn't change the value. What about negative exponents? If I had ( t^3 cdot t^{-4} ), would that be ( t^{3 + (-4)} = t^{-1} )? Yeah, that seems right. And ( t^{-1} ) is the same as ( frac{1}{t} ). So, it still works with negative exponents. Okay, going back to the original problem, ( t^3 cdot t^4 ). Since both terms have the same base ( t ), I can add the exponents. So, 3 plus 4 is 7, which gives me ( t^7 ). Let me see if I can think of a real-life example to make this more concrete. Suppose I have a bacteria culture that doubles every hour. If I start with one bacterium, after 3 hours, there would be ( 2^3 = 8 ) bacteria. After 4 more hours, it would double again four times, so ( 2^4 = 16 ). If I want to know how many bacteria there are after 3 hours and then 4 more hours, I could multiply ( 2^3 cdot 2^4 ). Using the same rule, that's ( 2^{3+4} = 2^7 = 128 ) bacteria. So, it works in that context too. Another example: if I'm calculating the area of a square, the formula is side length squared. If I have a square with side length ( t^3 ), the area would be ( (t^3)^2 = t^{6} ). Wait, that's a different exponent rule, though—when you raise a power to a power, you multiply the exponents. So, that's a separate rule, but it's good to remember these different properties. But in this case, we're just multiplying two terms with the same base, so adding exponents is the way to go. I think I've got it. The key takeaway is that when multiplying like bases with exponents, you add the exponents. So, ( t^3 cdot t^4 = t^{3+4} = t^7 ). Just to make sure I'm not missing anything, let me try plugging in a number for ( t ) and see if both the original expression and the simplified expression give the same result. Let's choose ( t = 2 ). Calculating ( t^3 cdot t^4 ) with ( t = 2 ):( 2^3 = 8 ) and ( 2^4 = 16 ). Multiplying them together: ( 8 cdot 16 = 128 ).Now, calculating ( t^7 ) with ( t = 2 ):( 2^7 = 128 ).They match! So, that confirms that ( t^3 cdot t^4 = t^7 ) is correct. What if ( t ) is a negative number? Let's try ( t = -2 ).Calculating ( (-2)^3 cdot (-2)^4 ):( (-2)^3 = -8 ) and ( (-2)^4 = 16 ). Multiplying them: ( -8 cdot 16 = -128 ).Now, calculating ( (-2)^7 ):( (-2)^7 = -128 ).Again, they match. So, even with a negative base, the rule holds true. What about if ( t ) is a fraction? Let's say ( t = frac{1}{2} ).Calculating ( left(frac{1}{2}right)^3 cdot left(frac{1}{2}right)^4 ):( left(frac{1}{2}right)^3 = frac{1}{8} ) and ( left(frac{1}{2}right)^4 = frac{1}{16} ). Multiplying them: ( frac{1}{8} cdot frac{1}{16} = frac{1}{128} ).Now, calculating ( left(frac{1}{2}right)^7 ):( left(frac{1}{2}right)^7 = frac{1}{128} ).They match again. So, regardless of whether ( t ) is positive, negative, or a fraction, the rule ( t^a cdot t^b = t^{a+b} ) holds true. I think I've covered different scenarios and checked the validity of the rule in each case. It seems consistent, so I'm confident that simplifying ( t^3 cdot t^4 ) to ( t^7 ) is correct. Just to recap, the steps are:1. Identify that both terms have the same base ( t ).2. Recall the exponent rule for multiplication: ( t^a cdot t^b = t^{a+b} ).3. Apply the rule by adding the exponents: ( 3 + 4 = 7 ).4. Write the simplified expression: ( t^7 ).5. Verify with different values of ( t ) to ensure the rule works universally.This process helps solidify the understanding of exponent rules and ensures that the simplification is accurate.

answer：First, I need to determine the total number of hockey games Tom attended this year and last year combined, which is 13 games.Tom attended 4 games this year and missed 7 games. This means there were a total of 11 games this year.To find out how many games Tom attended last year, I subtract the number of games he attended this year from the total number of games he attended in both years.So, 13 total games minus 4 games this year equals 9 games last year.Therefore, Tom attended 9 hockey games last year.

answer：Alright, so I have this problem about calculating the total volume of ice cream in a cone setup. Let me try to visualize it first. There's a right, circular cone that's 10 inches tall with a radius of 3 inches at the opening. The inside of the cone is completely filled with ice cream. On top of that, there's a hemisphere covering the opening of the cone, and then a cylindrical layer of ice cream on top of the hemisphere, which is 2 inches tall with the same radius as the cone's opening.Okay, so I need to find the total volume of ice cream. That means I have to calculate the volume of the cone, the hemisphere, and the cylinder separately and then add them all together. Let me recall the formulas for these volumes.First, the volume of a cone is given by ( V_{text{cone}} = frac{1}{3} pi r^2 h ), where ( r ) is the radius and ( h ) is the height. For the cone, the radius ( r ) is 3 inches, and the height ( h ) is 10 inches. Plugging those values in, I get:( V_{text{cone}} = frac{1}{3} pi (3)^2 (10) )Calculating that, ( 3^2 ) is 9, multiplied by 10 gives 90, and then divided by 3 is 30. So, the volume of the cone is ( 30pi ) cubic inches.Next, the hemisphere. A hemisphere is half of a sphere, so the volume formula is half of the sphere's volume. The volume of a sphere is ( frac{4}{3} pi r^3 ), so half of that is ( frac{2}{3} pi r^3 ). The radius of the hemisphere is the same as the cone's radius, which is 3 inches. Plugging that in:( V_{text{hemisphere}} = frac{2}{3} pi (3)^3 )Calculating that, ( 3^3 ) is 27, multiplied by ( frac{2}{3} ) gives 18. So, the volume of the hemisphere is ( 18pi ) cubic inches.Now, the cylindrical layer on top of the hemisphere. The volume of a cylinder is ( V_{text{cylinder}} = pi r^2 h ). The radius here is also 3 inches, and the height is 2 inches. Plugging in those values:( V_{text{cylinder}} = pi (3)^2 (2) )Calculating that, ( 3^2 ) is 9, multiplied by 2 gives 18. So, the volume of the cylinder is ( 18pi ) cubic inches.Alright, so now I have all three volumes:- Cone: ( 30pi )- Hemisphere: ( 18pi )- Cylinder: ( 18pi )To find the total volume of ice cream, I just need to add them all together:( V_{text{total}} = 30pi + 18pi + 18pi )Adding those up, 30 plus 18 is 48, plus another 18 is 66. So, the total volume is ( 66pi ) cubic inches.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For the cone: ( frac{1}{3} pi (3)^2 (10) ). Yes, ( 3^2 = 9 ), times 10 is 90, divided by 3 is 30. That seems right.For the hemisphere: ( frac{2}{3} pi (3)^3 ). ( 3^3 = 27 ), times ( frac{2}{3} ) is 18. That looks good.For the cylinder: ( pi (3)^2 (2) ). ( 3^2 = 9 ), times 2 is 18. Correct.Adding them up: 30 + 18 + 18. 30 + 18 is 48, plus 18 is 66. Yep, that's correct.So, I think I did everything right. The total volume of ice cream is ( 66pi ) cubic inches.

answer：Okay, so I've got this quadratic equation here: (x^2 - 6x + 5 = 2x - 8). Hmm, I need to find the sum of its solutions. Alright, let's start by understanding what the equation is saying.First, I remember that a quadratic equation is usually in the form (ax^2 + bx + c = 0). So, I think the first step is to rearrange this equation to match that standard form. That means I need to get all the terms on one side so that the other side is zero.Looking at the equation, on the left side, I have (x^2 - 6x + 5), and on the right side, there's (2x - 8). To move everything to the left, I can subtract (2x) and add (8) to both sides. Let's see:Starting with:[x^2 - 6x + 5 = 2x - 8]Subtract (2x) from both sides:[x^2 - 6x - 2x + 5 = -8]Simplify the left side:[x^2 - 8x + 5 = -8]Now, add (8) to both sides to get rid of the negative number on the right:[x^2 - 8x + 5 + 8 = 0]Simplify that:[x^2 - 8x + 13 = 0]Alright, now the equation is in the standard quadratic form (ax^2 + bx + c = 0), where (a = 1), (b = -8), and (c = 13).I remember there's a formula related to the sum of the roots of a quadratic equation. It goes like this: the sum of the roots is (-b/a). So, in this case, that would be (-(-8)/1), which simplifies to (8/1 = 8).Wait, let me make sure I'm not making a mistake here. Sometimes signs can be tricky. So, (b) is (-8), and when I plug it into the formula (-b/a), it becomes (-(-8)/1), which is indeed (8). That seems right.Just to double-check, maybe I can try solving the equation using the quadratic formula and see if the sum of the solutions matches. The quadratic formula is (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}).Plugging in the values:[x = frac{-(-8) pm sqrt{(-8)^2 - 4(1)(13)}}{2(1)}][x = frac{8 pm sqrt{64 - 52}}{2}][x = frac{8 pm sqrt{12}}{2}][x = frac{8 pm 2sqrt{3}}{2}][x = 4 pm sqrt{3}]So, the two solutions are (4 + sqrt{3}) and (4 - sqrt{3}). If I add them together:[(4 + sqrt{3}) + (4 - sqrt{3}) = 8]Yep, that matches the sum I got earlier using the formula. So, the sum of the solutions is definitely (8).I think I've got it. The key steps were rearranging the original equation into standard quadratic form and then applying the sum of roots formula. It's always good to verify by solving the equation fully, just to be sure there were no mistakes in the initial steps.

answer：First, I need to determine the circumference of the circular field because the fencing cost depends on the total length of the fence required.The formula for the circumference ( C ) of a circle is:[C = pi times text{diameter}]Given that the diameter is 16 meters, I can substitute this value into the formula:[C = pi times 16 , text{m}]Using the approximate value of ( pi ) as 3.14, the calculation becomes:[C = 3.14 times 16 , text{m} = 50.24 , text{m}]Now that I have the total length of the fence, I can calculate the cost by multiplying the length by the rate per meter:[text{Total cost} = 50.24 , text{m} times 3 , text{Rs/m} = 150.72 , text{Rs}]Therefore, the total cost of fencing around the circular field is approximately Rs. 150.72.