💡Okay, so I have this problem here where I need to convert a polar equation into a Cartesian equation. The polar equation given is ρ sin(θ - π/4) = √2. Hmm, I remember that converting between polar and Cartesian coordinates involves using some trigonometric identities and substitutions. Let me try to recall the steps.First, I know that in polar coordinates, ρ is the radius, and θ is the angle from the positive x-axis. To convert this to Cartesian coordinates, I should express ρ and θ in terms of x and y. The basic conversions are x = ρ cos θ and y = ρ sin θ. Maybe I can use these to rewrite the given equation.Looking at the equation ρ sin(θ - π/4) = √2, I think I can use the sine of a difference identity here. The identity is sin(A - B) = sin A cos B - cos A sin B. Applying this to sin(θ - π/4), it should become sin θ cos(π/4) - cos θ sin(π/4). Let me write that out:sin(θ - π/4) = sin θ cos(π/4) - cos θ sin(π/4)I remember that cos(π/4) and sin(π/4) are both √2/2. So substituting those in:sin(θ - π/4) = sin θ (√2/2) - cos θ (√2/2)So now, the original equation becomes:ρ [sin θ (√2/2) - cos θ (√2/2)] = √2Let me distribute the ρ:(ρ sin θ)(√2/2) - (ρ cos θ)(√2/2) = √2Now, I can substitute ρ sin θ with y and ρ cos θ with x, since x = ρ cos θ and y = ρ sin θ. So substituting those in:(y)(√2/2) - (x)(√2/2) = √2So that simplifies to:(√2/2)y - (√2/2)x = √2Hmm, I can factor out √2/2 from the left side:√2/2 (y - x) = √2Now, to solve for y - x, I can divide both sides by √2/2. Dividing by a fraction is the same as multiplying by its reciprocal, so:(y - x) = √2 / (√2/2)Simplify the right side:√2 divided by (√2/2) is equal to √2 * (2/√2) = 2So, y - x = 2If I rearrange this, it becomes:y = x + 2But usually, Cartesian equations are written in the form Ax + By + C = 0. So, subtracting y from both sides:x - y + 2 = 0Wait, let me check that again. If y - x = 2, then bringing x and y to the same side:y - x - 2 = 0Which is the same as:-x + y - 2 = 0But if I multiply both sides by -1 to make the x coefficient positive, it becomes:x - y + 2 = 0Yes, that looks correct. So, the Cartesian equation is x - y + 2 = 0.Let me just verify my steps to make sure I didn't make any mistakes. Starting from the polar equation, I used the sine difference identity correctly, substituted the known values for cos(π/4) and sin(π/4), then replaced ρ sin θ with y and ρ cos θ with x. After simplifying, I ended up with y - x = 2, which I rearranged into the standard linear form. Everything seems to check out.I think that's it. The Cartesian equation of the curve C is x - y + 2 = 0.**Final Answer**The Cartesian coordinate equation of (C) is boxed{x - y + 2 = 0}.

💡Okay, so I need to find the remainder when 2 raised to the 19th power is divided by 7. Hmm, that sounds like a modular arithmetic problem. I remember that modular arithmetic deals with remainders, so this should be the right approach.First, I think I should try to find a pattern or a cycle in the powers of 2 modulo 7. Maybe if I list out some powers of 2 and see what their remainders are when divided by 7, I can spot a repeating pattern.Let me start calculating:- 2^1 = 2. When I divide 2 by 7, the remainder is 2.- 2^2 = 4. Dividing 4 by 7 gives a remainder of 4.- 2^3 = 8. Dividing 8 by 7 gives a remainder of 1 because 7 goes into 8 once with a remainder of 1.- 2^4 = 16. Dividing 16 by 7, 7 times 2 is 14, so the remainder is 2.- 2^5 = 32. Dividing 32 by 7, 7 times 4 is 28, so the remainder is 4.- 2^6 = 64. Dividing 64 by 7, 7 times 9 is 63, so the remainder is 1.Wait a minute, I see a pattern here. The remainders are cycling through 2, 4, 1, and then repeating again. So, the cycle is 2, 4, 1, and then repeats every 3 exponents. That means every time the exponent increases by 3, the remainder cycles back to the start.So, the cycle length is 3. That means 2^1 ≡ 2 mod 7, 2^2 ≡ 4 mod 7, 2^3 ≡ 1 mod 7, and then 2^4 ≡ 2 mod 7 again, and so on.Now, since the cycle repeats every 3 exponents, I can use this to find 2^19 mod 7. I need to figure out where 19 falls in this cycle.To do that, I can divide 19 by the cycle length, which is 3, and find the remainder. That will tell me which position in the cycle 2^19 corresponds to.So, 19 divided by 3 is 6 with a remainder of 1. That means 19 = 3*6 + 1. Therefore, 2^19 is equivalent to 2^(3*6 + 1) = (2^3)^6 * 2^1.Since 2^3 ≡ 1 mod 7, then (2^3)^6 ≡ 1^6 ≡ 1 mod 7. So, 2^19 ≡ 1 * 2 ≡ 2 mod 7.Therefore, the remainder when 2^19 is divided by 7 is 2.Wait, let me double-check that. If the cycle is 2, 4, 1, then for exponents:1: 22: 43: 14: 25: 46: 17: 28: 49: 110: 211: 412: 113: 214: 415: 116: 217: 418: 119: 2Yes, that matches. So, 2^19 mod 7 is indeed 2.I think that makes sense. The key was recognizing the repeating cycle of remainders every 3 exponents. Once I identified that, I could break down the exponent 19 into multiples of 3 plus a remainder, which simplified the calculation.

💡Alright, so I have this problem here about a function and its derivative. Let me try to figure it out step by step. The function is given as f(x) = 2x³ + ax² + bx + 1. They mention its derivative, f’(x), and say that the graph of y = f’(x) is symmetric about the line x = -1/2. Also, they tell me that f’(1) = 0. I need to find the values of a and b first, and then find the extreme values of f(x).Okay, starting with part (I). I know that the derivative of f(x) will give me f’(x). So let me compute that. The derivative of 2x³ is 6x², the derivative of ax² is 2ax, the derivative of bx is b, and the derivative of 1 is 0. So putting it all together, f’(x) = 6x² + 2ax + b.Now, the graph of y = f’(x) is a quadratic function, right? Quadratic functions are parabolas, and they are symmetric about their vertex. The vertex form of a quadratic is y = A(x - h)² + k, where (h, k) is the vertex. The axis of symmetry is x = h. So in this case, the graph is symmetric about x = -1/2, which means the vertex is at x = -1/2.I remember that for a quadratic in standard form, y = Ax² + Bx + C, the axis of symmetry is given by x = -B/(2A). In our case, f’(x) = 6x² + 2ax + b, so A is 6, B is 2a. Therefore, the axis of symmetry is x = - (2a)/(2*6) = -a/6. They say this is equal to -1/2. So, setting -a/6 equal to -1/2.Let me write that equation: -a/6 = -1/2. To solve for a, I can multiply both sides by 6: -a = -3. Then, multiplying both sides by -1, I get a = 3. Okay, so a is 3. Got that.Next, they tell me that f’(1) = 0. So, plugging x = 1 into f’(x), which is 6x² + 2ax + b, and setting it equal to 0. Let's do that.f’(1) = 6*(1)² + 2*a*(1) + b = 6 + 2a + b = 0.We already found that a = 3, so plugging that in: 6 + 2*3 + b = 0. That simplifies to 6 + 6 + b = 0, so 12 + b = 0. Therefore, b = -12. Okay, so b is -12.So, part (I) gives us a = 3 and b = -12. That seems straightforward.Moving on to part (II), we need to find the extreme values of the function f(x). Extreme values occur where the derivative is zero or undefined, but since f’(x) is a polynomial, it's defined everywhere. So, we just need to find where f’(x) = 0.From part (I), we have f’(x) = 6x² + 6x - 12. Let me write that down: f’(x) = 6x² + 6x - 12. To find critical points, set this equal to zero.So, 6x² + 6x - 12 = 0. Let me factor out a 6 first: 6(x² + x - 2) = 0. So, x² + x - 2 = 0. Now, let's factor this quadratic. Looking for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1. So, (x + 2)(x - 1) = 0. Therefore, x = -2 or x = 1.So, the critical points are at x = -2 and x = 1. Now, to determine whether these are maxima or minima, I can use the second derivative test or analyze the sign changes of f’(x).Let me compute the second derivative, f''(x). The second derivative of f(x) is the derivative of f’(x). So, f’(x) = 6x² + 6x - 12, so f''(x) = 12x + 6.Now, evaluate f''(x) at the critical points.First, at x = -2: f''(-2) = 12*(-2) + 6 = -24 + 6 = -18. Since this is negative, the function is concave down at x = -2, so this is a local maximum.Next, at x = 1: f''(1) = 12*(1) + 6 = 12 + 6 = 18. This is positive, so the function is concave up at x = 1, meaning this is a local minimum.Therefore, f(x) has a local maximum at x = -2 and a local minimum at x = 1.Now, let's find the corresponding y-values for these x-values to get the extreme values.First, compute f(-2). f(x) = 2x³ + 3x² -12x + 1.f(-2) = 2*(-2)³ + 3*(-2)² -12*(-2) + 1.Calculating each term:2*(-2)³ = 2*(-8) = -163*(-2)² = 3*(4) = 12-12*(-2) = 24And the constant term is 1.So, adding them up: -16 + 12 + 24 + 1.Let's compute step by step:-16 + 12 = -4-4 + 24 = 2020 + 1 = 21So, f(-2) = 21. That's the local maximum.Next, compute f(1). f(1) = 2*(1)³ + 3*(1)² -12*(1) + 1.Calculating each term:2*(1)³ = 2*1 = 23*(1)² = 3*1 = 3-12*(1) = -12And the constant term is 1.Adding them up: 2 + 3 - 12 + 1.Compute step by step:2 + 3 = 55 - 12 = -7-7 + 1 = -6So, f(1) = -6. That's the local minimum.Therefore, the function f(x) has a local maximum of 21 at x = -2 and a local minimum of -6 at x = 1.Let me just double-check my calculations to make sure I didn't make any mistakes.For f(-2):2*(-8) = -163*(4) = 12-12*(-2) = 241 is 1-16 + 12 = -4-4 + 24 = 2020 + 1 = 21. That seems correct.For f(1):2*1 = 23*1 = 3-12*1 = -121 is 12 + 3 = 55 - 12 = -7-7 + 1 = -6. That also seems correct.Okay, so I think I did that right. So, the extreme values are 21 and -6 at x = -2 and x = 1, respectively.**Final Answer**(I) The values of ( a ) and ( b ) are ( boxed{3} ) and ( boxed{-12} ) respectively.(II) The extreme values of the function ( f(x) ) are a maximum of ( boxed{21} ) at ( x = -2 ) and a minimum of ( boxed{-6} ) at ( x = 1 ).

💡Alright, so I've got this problem about the Chinese zodiac signs and three students choosing gifts. Let me try to understand it step by step. First, the Chinese zodiac has twelve signs, and they're paired into six groups: rat and ox, tiger and rabbit, dragon and snake, horse and sheep, monkey and rooster, dog and pig. Each pair is considered complementary, making a perfect personality. So, there are six groups, each with two zodiac signs.Now, there are three students: A, B, and C. They choose a group as a gift in turn. That means A picks first, then B, then C. Each student has specific preferences:- Student A likes ox and horse.- Student B likes ox, dog, and sheep.- Student C likes all the mascots.The condition is that the gifts chosen by A, B, and C must all contain their favorite zodiac signs. So, we need to figure out how many different ways they can choose the gifts under these conditions.Let me break this down.First, let's list out the six groups:1. Rat and Ox2. Tiger and Rabbit3. Dragon and Snake4. Horse and Sheep5. Monkey and Rooster6. Dog and PigNow, Student A likes ox and horse. So, A can choose either the group with Rat and Ox or the group with Horse and Sheep because those are the only groups containing ox or horse.Similarly, Student B likes ox, dog, and sheep. So, B can choose the group with Rat and Ox, Dog and Pig, or Horse and Sheep.Student C likes all the mascots, so C can choose any remaining group.But since they choose in turn, A first, then B, then C, we need to consider the choices sequentially and ensure that each student gets a group that includes their favorite sign.Let me consider the possible choices for A first.**Case 1: A chooses Rat and Ox**If A chooses Rat and Ox, then the remaining groups are:1. Tiger and Rabbit2. Dragon and Snake3. Horse and Sheep4. Monkey and Rooster5. Dog and PigNow, B needs to choose a group that includes ox, dog, or sheep. But ox is already taken by A, so B can choose either Dog and Pig or Horse and Sheep.But wait, Horse and Sheep includes sheep, which B also likes. So, B has two choices:- Dog and Pig- Horse and SheepLet's explore both subcases.**Subcase 1a: B chooses Dog and Pig**If B chooses Dog and Pig, then the remaining groups are:1. Tiger and Rabbit2. Dragon and Snake3. Horse and Sheep4. Monkey and RoosterNow, C can choose any of these four groups since C likes all mascots. So, C has 4 choices.**Subcase 1b: B chooses Horse and Sheep**If B chooses Horse and Sheep, then the remaining groups are:1. Tiger and Rabbit2. Dragon and Snake3. Dog and Pig4. Monkey and RoosterAgain, C can choose any of these four groups, so C has 4 choices.Therefore, in Case 1, where A chooses Rat and Ox, B has 2 choices, and for each of B's choices, C has 4 choices. So, the total number of ways in Case 1 is 2 * 4 = 8 ways.**Case 2: A chooses Horse and Sheep**Now, let's consider the other possibility where A chooses Horse and Sheep. Then, the remaining groups are:1. Rat and Ox2. Tiger and Rabbit3. Dragon and Snake4. Monkey and Rooster5. Dog and PigAgain, B needs to choose a group that includes ox, dog, or sheep. Ox is in Rat and Ox, dog is in Dog and Pig, and sheep is already taken by A. So, B can choose either Rat and Ox or Dog and Pig.Let's explore these subcases.**Subcase 2a: B chooses Rat and Ox**If B chooses Rat and Ox, then the remaining groups are:1. Tiger and Rabbit2. Dragon and Snake3. Monkey and Rooster4. Dog and PigC can choose any of these four groups, so C has 4 choices.**Subcase 2b: B chooses Dog and Pig**If B chooses Dog and Pig, then the remaining groups are:1. Rat and Ox2. Tiger and Rabbit3. Dragon and Snake4. Monkey and RoosterAgain, C has 4 choices.Therefore, in Case 2, where A chooses Horse and Sheep, B has 2 choices, and for each of B's choices, C has 4 choices. So, the total number of ways in Case 2 is also 2 * 4 = 8 ways.**Total Number of Ways**Now, to find the total number of different ways, we add the number of ways from both cases:Case 1: 8 waysCase 2: 8 waysTotal: 8 + 8 = 16 waysSo, the total number of different ways to choose the gifts, ensuring all students get their favorite zodiac signs, is 16 ways.

💡Alright, so I have this function ( f(x) = ln x + frac{a}{x+1} ) where ( a ) is a real number. There are three parts to this problem, and I need to tackle each one step by step.Starting with part (I): When ( a = frac{9}{2} ), the function becomes ( f(x) = ln x + frac{9}{2(x+1)} ). They want me to find the range of ( k ) such that ( g(x) = f(x) - k ) has only one zero point. Hmm, okay. So, essentially, I need to find the values of ( k ) where the equation ( f(x) = k ) has only one solution.First, I think I need to analyze the behavior of ( f(x) ). Since ( f(x) ) involves a logarithm and a rational function, I should consider its domain, which is ( x > 0 ) because of the ( ln x ) term.To find the number of solutions, I should look at the critical points of ( f(x) ) by taking its derivative. Let's compute ( f'(x) ):( f'(x) = frac{1}{x} - frac{9}{2(x+1)^2} ).Simplify this:( f'(x) = frac{1}{x} - frac{9}{2(x+1)^2} ).To find critical points, set ( f'(x) = 0 ):( frac{1}{x} = frac{9}{2(x+1)^2} ).Multiply both sides by ( 2x(x+1)^2 ) to eliminate denominators:( 2(x+1)^2 = 9x ).Expand ( (x+1)^2 ):( 2(x^2 + 2x + 1) = 9x ).Which simplifies to:( 2x^2 + 4x + 2 = 9x ).Bring all terms to one side:( 2x^2 - 5x + 2 = 0 ).Factor the quadratic:Looking for two numbers that multiply to ( 4 ) (since ( 2 times 2 )) and add to ( -5 ). Hmm, that would be ( -4 ) and ( -1 ).So, ( 2x^2 - 4x - x + 2 = 0 ).Factor by grouping:( 2x(x - 2) -1(x - 2) = 0 ).Thus, ( (2x - 1)(x - 2) = 0 ).So, critical points at ( x = frac{1}{2} ) and ( x = 2 ).Now, I need to determine the behavior of ( f(x) ) around these critical points and at the boundaries.First, let's compute the limits as ( x ) approaches 0 and infinity.As ( x to 0^+ ), ( ln x to -infty ) and ( frac{9}{2(x+1)} to frac{9}{2} ). So, overall, ( f(x) to -infty ).As ( x to infty ), ( ln x to infty ) and ( frac{9}{2(x+1)} to 0 ). So, ( f(x) to infty ).Now, let's evaluate ( f(x) ) at the critical points.At ( x = frac{1}{2} ):( fleft(frac{1}{2}right) = ln left(frac{1}{2}right) + frac{9}{2left(frac{1}{2} + 1right)} ).Simplify:( ln left(frac{1}{2}right) = -ln 2 ).( frac{9}{2 times frac{3}{2}} = frac{9}{3} = 3 ).So, ( fleft(frac{1}{2}right) = 3 - ln 2 ).At ( x = 2 ):( f(2) = ln 2 + frac{9}{2(2 + 1)} = ln 2 + frac{9}{6} = ln 2 + frac{3}{2} ).So, ( f(2) = frac{3}{2} + ln 2 ).Now, I need to determine the behavior of ( f(x) ) around these points.Since ( f'(x) ) changes sign from positive to negative at ( x = frac{1}{2} ) and from negative to positive at ( x = 2 ), ( x = frac{1}{2} ) is a local maximum, and ( x = 2 ) is a local minimum.So, the function increases from ( -infty ) to ( x = frac{1}{2} ), reaching a maximum of ( 3 - ln 2 ), then decreases until ( x = 2 ), reaching a minimum of ( frac{3}{2} + ln 2 ), and then increases again towards ( infty ).Therefore, the graph of ( f(x) ) has a "hill" from 0 to ( frac{1}{2} ), a "valley" from ( frac{1}{2} ) to 2, and then another "hill" beyond 2.Now, the function ( g(x) = f(x) - k ) will have only one zero when the horizontal line ( y = k ) intersects the graph of ( f(x) ) exactly once.Looking at the behavior:- For ( k > 3 - ln 2 ), the line ( y = k ) will intersect the rightmost "hill" only once because ( f(x) ) goes to infinity as ( x ) increases.- For ( k < frac{3}{2} + ln 2 ), the line ( y = k ) will intersect the leftmost part of the graph only once because ( f(x) ) goes to negative infinity as ( x ) approaches 0.However, between ( frac{3}{2} + ln 2 ) and ( 3 - ln 2 ), the line ( y = k ) will intersect the graph three times: once on the left of ( frac{1}{2} ), once between ( frac{1}{2} ) and 2, and once after 2.Therefore, to have only one zero, ( k ) must be either greater than ( 3 - ln 2 ) or less than ( frac{3}{2} + ln 2 ).So, the range of ( k ) is ( k > 3 - ln 2 ) or ( k < frac{3}{2} + ln 2 ).Moving on to part (II): When ( a = 2 ), compare the size of ( f(x) ) with 1.So, ( f(x) = ln x + frac{2}{x + 1} ). We need to determine when ( f(x) > 1 ), ( f(x) = 1 ), or ( f(x) < 1 ).Again, let's consider the function ( h(x) = f(x) - 1 = ln x + frac{2}{x + 1} - 1 ). We can analyze the behavior of ( h(x) ).First, compute the derivative ( h'(x) ):( h'(x) = frac{1}{x} - frac{2}{(x + 1)^2} ).Simplify:( h'(x) = frac{1}{x} - frac{2}{(x + 1)^2} ).To analyze the monotonicity, let's see if ( h'(x) ) is always positive or negative.Let me compute ( h'(x) ):Let me find a common denominator:( h'(x) = frac{(x + 1)^2 - 2x}{x(x + 1)^2} ).Expand numerator:( (x + 1)^2 - 2x = x^2 + 2x + 1 - 2x = x^2 + 1 ).So, ( h'(x) = frac{x^2 + 1}{x(x + 1)^2} ).Since ( x > 0 ), the denominator is positive, and the numerator ( x^2 + 1 ) is always positive. Therefore, ( h'(x) > 0 ) for all ( x > 0 ).This means ( h(x) ) is strictly increasing on its domain.Now, let's evaluate ( h(x) ) at ( x = 1 ):( h(1) = ln 1 + frac{2}{2} - 1 = 0 + 1 - 1 = 0 ).Since ( h(x) ) is strictly increasing, for ( x > 1 ), ( h(x) > 0 ), which implies ( f(x) > 1 ).For ( x < 1 ), ( h(x) < 0 ), which implies ( f(x) < 1 ).At ( x = 1 ), ( h(x) = 0 ), so ( f(x) = 1 ).Therefore, the comparison is:- ( f(x) < 1 ) when ( 0 < x < 1 )- ( f(x) = 1 ) when ( x = 1 )- ( f(x) > 1 ) when ( x > 1 )Finally, part (III): Prove that ( ln(n + 1) > frac{1}{3} + frac{1}{5} + frac{1}{7} + ldots + frac{1}{2n + 1} ) where ( n in mathbb{N}^* ).Hmm, this seems like it might relate to the integral of ( 1/x ) or perhaps using the result from part (II).Wait, in part (II), we saw that for ( x > 1 ), ( ln x + frac{2}{x + 1} > 1 ). Maybe we can manipulate this inequality.Let me rearrange the inequality from part (II):( ln x > 1 - frac{2}{x + 1} ).Simplify the right side:( 1 - frac{2}{x + 1} = frac{(x + 1) - 2}{x + 1} = frac{x - 1}{x + 1} ).So, ( ln x > frac{x - 1}{x + 1} ) for ( x > 1 ).Interesting. So, for each ( x > 1 ), ( ln x > frac{x - 1}{x + 1} ).Perhaps I can use this inequality for specific values of ( x ) to sum up the series.Let me consider ( x = frac{k + 1}{k} ) for ( k = 1, 2, ldots, n ).So, substituting ( x = frac{k + 1}{k} ), we have:( lnleft(frac{k + 1}{k}right) > frac{frac{k + 1}{k} - 1}{frac{k + 1}{k} + 1} ).Simplify the right side:Numerator: ( frac{k + 1}{k} - 1 = frac{1}{k} ).Denominator: ( frac{k + 1}{k} + 1 = frac{k + 1 + k}{k} = frac{2k + 1}{k} ).So, the fraction becomes ( frac{frac{1}{k}}{frac{2k + 1}{k}} = frac{1}{2k + 1} ).Therefore, ( lnleft(frac{k + 1}{k}right) > frac{1}{2k + 1} ).Now, sum both sides from ( k = 1 ) to ( k = n ):Left side: ( sum_{k=1}^{n} lnleft(frac{k + 1}{k}right) = ln(n + 1) ) because it's a telescoping series.Right side: ( sum_{k=1}^{n} frac{1}{2k + 1} ).Therefore, ( ln(n + 1) > sum_{k=1}^{n} frac{1}{2k + 1} ).Which is exactly what we needed to prove.So, putting it all together, by using the inequality from part (II) and substituting specific values, we can establish the desired inequality.**Final Answer**(I) The range of ( k ) is ( boxed{(-infty, frac{3}{2} + ln 2) cup (3 - ln 2, +infty)} ).(II) ( f(x) < 1 ) when ( 0 < x < 1 ), ( f(x) = 1 ) when ( x = 1 ), and ( f(x) > 1 ) when ( x > 1 ).(III) The inequality is proven as ( boxed{ln(n + 1) > frac{1}{3} + frac{1}{5} + frac{1}{7} + ldots + frac{1}{2n + 1}} ).

💡Alright, so I've got these four math problems to solve. Let me take them one by one and see if I can figure them out. I'm going to write down my thoughts as I go along, even if I make some mistakes or get confused. Hopefully, by the end, I'll have the right answers.**Problem 1: Calculate -7 + (-3) - 4 - |-8|**Okay, starting with the first problem. It looks like a series of additions and subtractions, and there's an absolute value involved. Let me break it down step by step.First, I see -7 plus (-3). Adding two negative numbers is like adding their magnitudes and keeping the negative sign. So, -7 plus -3 should be -10. That part seems straightforward.Next, we have -10 minus 4. Subtracting 4 from -10 is like moving further into the negatives. So, -10 minus 4 would be -14.Now, the last part is -|-8|. Absolute value of -8 is 8, right? Because absolute value makes any number inside it positive. So, -|-8| is the same as -8. So, now I have -14 minus 8.Wait, hold on. Is that right? So, -14 minus 8 is like subtracting a positive number, which is the same as adding a negative. So, -14 - 8 is -22.Let me double-check that. Starting from -7, adding -3 gives -10. Then subtracting 4 gives -14. Then subtracting 8 (because of the absolute value) gives -22. Yeah, that seems correct.**Problem 2: Calculate (1/2 - 5/9 + 7/12) × (-36)**Alright, moving on to the second problem. This one involves fractions and then multiplying by a negative number. Hmm, fractions can be tricky, but let's see.First, I need to compute the expression inside the parentheses: 1/2 minus 5/9 plus 7/12. To add or subtract fractions, they need a common denominator. Let's find the least common denominator (LCD) for 2, 9, and 12.The denominators are 2, 9, and 12. Let's list the multiples:- Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24...- Multiples of 9: 9, 18, 27, 36...- Multiples of 12: 12, 24, 36...The smallest common multiple is 36. So, the LCD is 36.Now, let's convert each fraction to have 36 as the denominator.1/2 = 18/365/9 = 20/367/12 = 21/36So, substituting back into the expression:18/36 - 20/36 + 21/36Now, let's perform the operations:18 - 20 + 21 = (18 - 20) + 21 = (-2) + 21 = 19So, the expression inside the parentheses simplifies to 19/36.Now, we need to multiply this by (-36):(19/36) × (-36) = ?Multiplying fractions by whole numbers: multiply the numerator by the whole number and keep the denominator.So, 19 × (-36) = -684Then, divide by 36: -684 / 36 = -19Wait, that seems straightforward. So, the result is -19.Let me verify:1/2 is 0.5, 5/9 is approximately 0.555..., and 7/12 is approximately 0.583...So, 0.5 - 0.555 + 0.583 ≈ 0.5 - 0.555 = -0.055 + 0.583 ≈ 0.528Then, 0.528 × (-36) ≈ -19.008Which is approximately -19, so that checks out.**Problem 3: Calculate -3² + 16 ÷ (-2) × 1/2 - (-1)²⁰²³**Okay, this one looks a bit more complicated with exponents and order of operations. Let's take it step by step.First, let's rewrite the expression:-3² + 16 ÷ (-2) × 1/2 - (-1)^2023I need to remember the order of operations: PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).So, first, handle the exponents.-3²: Wait, is this -(3²) or (-3)²? The way it's written, it's -3 squared, which is -(3²) = -9.Next, (-1)^2023: Since the exponent is odd, (-1) raised to any odd power is -1.So, substituting back:-9 + 16 ÷ (-2) × 1/2 - (-1)Simplify the expression:-9 + (16 ÷ (-2)) × (1/2) - (-1)Now, handle the division and multiplication from left to right.First, 16 ÷ (-2) = -8Then, -8 × (1/2) = -4So, now the expression is:-9 + (-4) - (-1)Simplify the additions and subtractions:-9 - 4 + 1Which is:(-9 - 4) + 1 = -13 + 1 = -12Wait, but I think I might have made a mistake here. Let me check again.Original expression:-3² + 16 ÷ (-2) × 1/2 - (-1)^2023Breaking it down:-3² = -916 ÷ (-2) = -8-8 × 1/2 = -4(-1)^2023 = -1So, substituting back:-9 + (-4) - (-1) = -9 - 4 + 1 = -12Yes, that seems correct.**Problem 4: Simplify 3a² - 2ab - a² + 5ab**Alright, the last problem is about simplifying an algebraic expression by combining like terms.The expression is: 3a² - 2ab - a² + 5abFirst, let's identify like terms.Terms with a²: 3a² and -a²Terms with ab: -2ab and 5abCombine the a² terms:3a² - a² = 2a²Combine the ab terms:-2ab + 5ab = 3abSo, putting it all together:2a² + 3abThat's the simplified form.Let me double-check:3a² - a² = 2a²-2ab + 5ab = 3abYes, that's correct.**Summary of Answers:**1. -222. -193. -124. 2a² + 3abI think I've got all the problems solved correctly. Let me just quickly go through each one again to make sure I didn't make any calculation errors.For Problem 1: -7 -3 -4 -8 = -22. Yep.Problem 2: (1/2 -5/9 +7/12) × (-36) = -19. That matches.Problem 3: -9 -4 +1 = -12. Correct.Problem 4: 3a² -a² = 2a²; -2ab +5ab = 3ab. So, 2a² +3ab. Right.Okay, I feel confident about these answers now.