💡Alright, so I've got this problem in front of me, and it's asking me to figure out which of the given expressions is correct. The options are labeled A through D, and each one involves some exponent rules. Hmm, exponent rules... I remember learning about those in algebra. Let me see if I can recall them correctly.First, let's go through each option one by one. Maybe if I break them down, it'll be easier to understand which one is right.Starting with option A: (a^{3} - a^{2} = a). Okay, so this is saying that if you subtract (a^{2}) from (a^{3}), you get (a). Hmm, does that make sense? Let me think about it. If I have (a^{3}), that's like (a) multiplied by itself three times, and (a^{2}) is (a) multiplied by itself two times. Subtracting them... I don't think they can be directly subtracted like that because they're not like terms. Like terms have the same exponent, right? So (a^{3}) and (a^{2}) are not like terms, so you can't combine them by subtraction to get (a). That doesn't seem right. So, I think option A is incorrect.Moving on to option B: ((a^{2})^{3} = a^{6}). Alright, this is about raising a power to another power. I remember that when you have an exponent raised to another exponent, you multiply the exponents. So, ((a^{2})^{3}) should be (a^{2 times 3}), which is (a^{6}). That matches what's given here. So, option B looks correct. But just to be thorough, let me check the other options too.Option C: (a^{3} cdot a^{2} = a^{6}). This is about multiplying two exponents with the same base. The rule here is that when you multiply like bases, you add the exponents. So, (a^{3} cdot a^{2}) should be (a^{3 + 2}), which is (a^{5}), not (a^{6}). So, option C is incorrect because it claims the result is (a^{6}) when it should be (a^{5}).Lastly, option D: ((a^{2})^{3} = a^{5}). Wait a minute, this looks similar to option B. Earlier, I figured out that ((a^{2})^{3}) equals (a^{6}). But here, it's saying it equals (a^{5}). That doesn't make sense because, as I calculated before, multiplying the exponents gives (a^{6}), not (a^{5}). So, option D is also incorrect.So, after going through all the options, it seems like only option B is correct. It properly applies the exponent rule for raising a power to another power by multiplying the exponents. The other options either incorrectly combine unlike terms, misapply the multiplication rule for exponents, or incorrectly state the result of raising a power to another power.I think I've got this figured out. Option B is the correct expression.

💡First, I need to understand that the incorrect result of multiplying a number ( a ) by 153 is 102325, and both of the digit '2's in this result are wrong.To find the correct result, I'll start by dividing the incorrect result by 153 to estimate the original number ( a ):[a approx frac{102325}{153} approx 668.79]Since ( a ) must be an integer, I'll round this to the nearest whole number, which is 669.Next, I'll multiply this estimated value of ( a ) by 153 to find the correct result:[669 times 153 = 102357]Finally, I'll verify that 102357 is indeed a multiple of 153:[102357 div 153 = 669]Since the division results in an integer, 102357 is the correct result of multiplying ( a ) by 153.

💡Okay, so I have this problem where set A is {1,2,3,4} and set B is {1,2}. The question is asking how many different sets C can satisfy the condition that when you take the union of B and C, you get A. Hmm, let me think about this step by step.First, I remember that the union of two sets includes all the elements that are in either of the sets. So, B union C should include everything that's in B and everything that's in C. Since B is {1,2}, and A is {1,2,3,4}, that means C has to provide the elements 3 and 4 because B doesn't have them. So, C must at least contain 3 and 4.But wait, can C have more elements than just 3 and 4? Well, if C has elements that are already in B, like 1 or 2, that's okay because the union would still just be A. So, C can have 3 and 4, and it can also have 1 and/or 2 as well. Let me try to list out all the possible sets C that satisfy this condition. Since C must include 3 and 4, the other elements (1 and 2) can be either included or not. So, for each of these elements, there are two choices: include it or not. That means for each of the elements 1 and 2, we have two options, which gives us 2 x 2 = 4 possible combinations.Let me write them out to make sure I haven't missed anything:1. C = {3,4}2. C = {3,4,1}3. C = {3,4,2}4. C = {3,4,1,2}Yep, that seems to cover all the possibilities. Each of these sets, when unioned with B, will give us A. For example, if C is {3,4}, then B union C is {1,2} union {3,4} = {1,2,3,4} which is A. Similarly, if C is {3,4,1}, then B union C is still {1,2,3,4} because B already has 1 and 2.I think that's all the possible sets. There are four of them, so the answer should be 4.

💡Okay, so I have this problem that I need to prove: the sum of certain binomial coefficients equals a Fibonacci number. Specifically, it says that ( C_{n}^{0} + C_{n-1}^{1} + C_{n-2}^{2} + ldots = F_{n+1} ). The hint mentions that this sum can be interpreted as the sum of elements in one diagonal of Pascal's Triangle. Hmm, interesting.First, I should recall what binomial coefficients are. ( C_{n}^{k} ) represents the number of ways to choose ( k ) elements from a set of ( n ) elements. In Pascal's Triangle, each row corresponds to the coefficients of the binomial expansion, so row ( n ) has entries ( C_{n}^{0}, C_{n}^{1}, ldots, C_{n}^{n} ).Now, the sum given is ( C_{n}^{0} + C_{n-1}^{1} + C_{n-2}^{2} + ldots ). If I visualize Pascal's Triangle, this seems like moving diagonally from the top right to the bottom left. For example, if ( n = 4 ), the sum would be ( C_{4}^{0} + C_{3}^{1} + C_{2}^{2} ). Let me compute that: ( 1 + 3 + 1 = 5 ). And ( F_{5} ) is indeed 5, since the Fibonacci sequence goes 1, 1, 2, 3, 5, 8, etc. So that checks out for ( n = 4 ).Let me try another example to see if this holds. Take ( n = 3 ). Then the sum is ( C_{3}^{0} + C_{2}^{1} + C_{1}^{2} ). Calculating each term: ( 1 + 2 + 1 = 4 ). And ( F_{4} ) is 3, wait, that doesn't match. Hmm, maybe I made a mistake. Wait, ( F_{4} ) is actually 3, but my sum is 4. That doesn't match. Did I compute the sum correctly?Wait, for ( n = 3 ), the sum should be ( C_{3}^{0} + C_{2}^{1} + C_{1}^{2} ). Let me double-check: ( C_{3}^{0} = 1 ), ( C_{2}^{1} = 2 ), ( C_{1}^{2} ) is actually 0 because you can't choose 2 elements from 1. So maybe the sum is ( 1 + 2 + 0 = 3 ), which matches ( F_{4} = 3 ). Okay, so I guess the sum stops when the lower index exceeds the upper index, which makes sense.So, for ( n = 3 ), the sum is 3, which is ( F_{4} ). For ( n = 4 ), the sum is 5, which is ( F_{5} ). Let me try ( n = 2 ): ( C_{2}^{0} + C_{1}^{1} + C_{0}^{2} ). That's ( 1 + 1 + 0 = 2 ), which is ( F_{3} = 2 ). Okay, that works too.So, the pattern seems to hold for these small values. Now, how can I prove this in general? Maybe I can use mathematical induction. Induction is often useful for proving statements that involve integers like this.Let me recall how induction works. I need to show that the statement is true for a base case, usually ( n = 0 ) or ( n = 1 ), and then assume it's true for some arbitrary ( n = k ) and prove it for ( n = k + 1 ).First, let's establish the base case. Let's take ( n = 0 ). Then the sum is ( C_{0}^{0} = 1 ). And ( F_{1} = 1 ). So, ( 1 = 1 ), which holds true.Next, let's take ( n = 1 ). The sum is ( C_{1}^{0} = 1 ). And ( F_{2} = 1 ). So, ( 1 = 1 ), which also holds.Now, let's assume that the statement is true for ( n = k ) and ( n = k - 1 ). That is, assume:1. ( C_{k}^{0} + C_{k-1}^{1} + C_{k-2}^{2} + ldots = F_{k+1} )2. ( C_{k-1}^{0} + C_{k-2}^{1} + C_{k-3}^{2} + ldots = F_{k} )Now, we need to show that the statement holds for ( n = k + 1 ). That is, we need to show:( C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + ldots = F_{k+2} )Let me write out the sum for ( n = k + 1 ):( S_{k+1} = C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + C_{k-2}^{3} + ldots )I can try to express this sum in terms of the sums for ( n = k ) and ( n = k - 1 ). Let's see:Notice that ( C_{k+1}^{0} = 1 ) and ( C_{k}^{1} = k ). But I'm not sure if that helps directly. Maybe I can use Pascal's identity, which states that ( C_{n}^{k} = C_{n-1}^{k-1} + C_{n-1}^{k} ).Let me try to apply Pascal's identity to each term in the sum ( S_{k+1} ). For example, ( C_{k}^{1} = C_{k-1}^{0} + C_{k-1}^{1} ). Similarly, ( C_{k-1}^{2} = C_{k-2}^{1} + C_{k-2}^{2} ), and so on.So, substituting these into ( S_{k+1} ):( S_{k+1} = C_{k+1}^{0} + (C_{k-1}^{0} + C_{k-1}^{1}) + (C_{k-2}^{1} + C_{k-2}^{2}) + ldots )Now, let's rearrange the terms:( S_{k+1} = C_{k+1}^{0} + C_{k-1}^{0} + C_{k-1}^{1} + C_{k-2}^{1} + C_{k-2}^{2} + ldots )Notice that ( C_{k+1}^{0} = 1 ) and ( C_{k-1}^{0} = 1 ). So, combining these:( S_{k+1} = 1 + 1 + (C_{k-1}^{1} + C_{k-2}^{1}) + (C_{k-2}^{2} + ldots) )Wait, this seems a bit messy. Maybe I can group the terms differently. Let's see:( S_{k+1} = C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + ldots )But ( C_{k}^{1} = C_{k-1}^{0} + C_{k-1}^{1} ), so substituting:( S_{k+1} = C_{k+1}^{0} + (C_{k-1}^{0} + C_{k-1}^{1}) + C_{k-1}^{2} + ldots )Now, ( C_{k+1}^{0} = 1 ) and ( C_{k-1}^{0} = 1 ), so:( S_{k+1} = 1 + 1 + C_{k-1}^{1} + C_{k-1}^{2} + ldots )Wait, but ( C_{k-1}^{1} + C_{k-1}^{2} + ldots ) is almost the sum ( S_{k-1} ), except ( S_{k-1} ) starts with ( C_{k-1}^{0} ). So, ( C_{k-1}^{1} + C_{k-1}^{2} + ldots = S_{k-1} - C_{k-1}^{0} = S_{k-1} - 1 ).Therefore, substituting back:( S_{k+1} = 1 + 1 + (S_{k-1} - 1) = 1 + S_{k-1} )But from our induction hypothesis, ( S_{k-1} = F_{k} ). So:( S_{k+1} = 1 + F_{k} )Wait, but we need to show that ( S_{k+1} = F_{k+2} ). Hmm, this seems off. Maybe I made a mistake in the substitution.Let me try another approach. Let's consider the sum ( S_{k+1} ) and see if it can be expressed in terms of ( S_{k} ) and ( S_{k-1} ).From the definition:( S_{k+1} = C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + ldots )Now, ( C_{k+1}^{0} = 1 ), and the rest of the terms ( C_{k}^{1} + C_{k-1}^{2} + ldots ) look similar to ( S_{k} ), but starting from ( C_{k}^{1} ) instead of ( C_{k}^{0} ).Wait, ( S_{k} = C_{k}^{0} + C_{k-1}^{1} + C_{k-2}^{2} + ldots ). So, if I subtract ( C_{k}^{0} ) from ( S_{k} ), I get ( C_{k-1}^{1} + C_{k-2}^{2} + ldots ), which is exactly the part after ( C_{k}^{1} ) in ( S_{k+1} ).But in ( S_{k+1} ), after ( C_{k+1}^{0} ), the next term is ( C_{k}^{1} ), which is different from ( C_{k-1}^{1} ). Hmm, maybe I need to adjust this.Alternatively, let's try to express ( S_{k+1} ) as ( C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + ldots ). Notice that ( C_{k}^{1} = C_{k-1}^{0} + C_{k-1}^{1} ) by Pascal's identity. Similarly, ( C_{k-1}^{2} = C_{k-2}^{1} + C_{k-2}^{2} ), and so on.So, substituting these into ( S_{k+1} ):( S_{k+1} = C_{k+1}^{0} + (C_{k-1}^{0} + C_{k-1}^{1}) + (C_{k-2}^{1} + C_{k-2}^{2}) + ldots )Now, let's rearrange the terms:( S_{k+1} = C_{k+1}^{0} + C_{k-1}^{0} + (C_{k-1}^{1} + C_{k-2}^{1}) + (C_{k-2}^{2} + ldots) )Notice that ( C_{k+1}^{0} = 1 ) and ( C_{k-1}^{0} = 1 ), so:( S_{k+1} = 1 + 1 + (C_{k-1}^{1} + C_{k-2}^{1}) + (C_{k-2}^{2} + ldots) )Now, ( C_{k-1}^{1} + C_{k-2}^{1} = C_{k}^{1} ) by Pascal's identity. Similarly, ( C_{k-2}^{2} + C_{k-3}^{2} = C_{k-1}^{2} ), and so on.Wait, but in our sum, after ( C_{k-1}^{1} + C_{k-2}^{1} ), we have ( C_{k-2}^{2} + ldots ). So, it seems like we can pair terms to form the sum ( S_{k} ).Let me see:( S_{k+1} = 1 + 1 + (C_{k}^{1}) + (C_{k-1}^{2}) + ldots )But ( C_{k}^{1} + C_{k-1}^{2} + ldots ) is exactly ( S_{k} - C_{k}^{0} ), since ( S_{k} = C_{k}^{0} + C_{k-1}^{1} + C_{k-2}^{2} + ldots ).So, ( S_{k+1} = 1 + 1 + (S_{k} - C_{k}^{0}) )Since ( C_{k}^{0} = 1 ), this simplifies to:( S_{k+1} = 1 + 1 + S_{k} - 1 = 1 + S_{k} )But from our induction hypothesis, ( S_{k} = F_{k+1} ). Therefore:( S_{k+1} = 1 + F_{k+1} )Wait, but we need to show that ( S_{k+1} = F_{k+2} ). Since the Fibonacci sequence satisfies ( F_{k+2} = F_{k+1} + F_{k} ), and from our earlier assumption, ( S_{k} = F_{k+1} ) and ( S_{k-1} = F_{k} ), then:( F_{k+2} = F_{k+1} + F_{k} = S_{k} + S_{k-1} )But from our previous step, ( S_{k+1} = 1 + S_{k} ). This doesn't directly match ( F_{k+2} = S_{k} + S_{k-1} ). Hmm, seems like there's a discrepancy here.Maybe I need to reconsider how I'm breaking down the sum. Let's go back to the expression:( S_{k+1} = C_{k+1}^{0} + C_{k}^{1} + C_{k-1}^{2} + ldots )Using Pascal's identity on each term:- ( C_{k}^{1} = C_{k-1}^{0} + C_{k-1}^{1} )- ( C_{k-1}^{2} = C_{k-2}^{1} + C_{k-2}^{2} )- ( C_{k-2}^{3} = C_{k-3}^{2} + C_{k-3}^{3} )- And so on.So, substituting these into ( S_{k+1} ):( S_{k+1} = C_{k+1}^{0} + (C_{k-1}^{0} + C_{k-1}^{1}) + (C_{k-2}^{1} + C_{k-2}^{2}) + (C_{k-3}^{2} + C_{k-3}^{3}) + ldots )Now, let's group the terms:- The first term is ( C_{k+1}^{0} = 1 )- The next terms are ( C_{k-1}^{0} + C_{k-2}^{1} + C_{k-3}^{2} + ldots )- And the remaining terms are ( C_{k-1}^{1} + C_{k-2}^{2} + C_{k-3}^{3} + ldots )Notice that ( C_{k-1}^{0} + C_{k-2}^{1} + C_{k-3}^{2} + ldots ) is exactly ( S_{k-1} ), and ( C_{k-1}^{1} + C_{k-2}^{2} + C_{k-3}^{3} + ldots ) is exactly ( S_{k} - C_{k}^{0} ), since ( S_{k} = C_{k}^{0} + C_{k-1}^{1} + C_{k-2}^{2} + ldots ).Therefore, we can write:( S_{k+1} = 1 + S_{k-1} + (S_{k} - 1) )Simplifying:( S_{k+1} = 1 + S_{k-1} + S_{k} - 1 = S_{k} + S_{k-1} )From our induction hypothesis, ( S_{k} = F_{k+1} ) and ( S_{k-1} = F_{k} ). Therefore:( S_{k+1} = F_{k+1} + F_{k} = F_{k+2} )Which is exactly what we needed to prove! So, by mathematical induction, the statement holds for all non-negative integers ( n ).I think this makes sense now. By breaking down each term using Pascal's identity and recognizing the structure of the sums, we can relate ( S_{k+1} ) to ( S_{k} ) and ( S_{k-1} ), which allows us to use the induction hypothesis to complete the proof.

💡First, I need to simplify each of the given roots individually.Starting with the cube root of 27, I recognize that 27 is equal to 3 raised to the power of 3. Therefore, the cube root of 27 simplifies to 3.Next, I look at the fourth root of 81. Since 81 is 3 raised to the power of 4, the fourth root of 81 also simplifies to 3.Finally, I simplify the square root of 25. Knowing that 25 is 5 squared, the square root of 25 simplifies to 5.Now that I have each term simplified to 3, 3, and 5 respectively, I can multiply these values together. Multiplying 3 by 3 gives me 9, and then multiplying 9 by 5 results in 45.Therefore, the product of the cube root of 27, the fourth root of 81, and the square root of 25 is 45.

💡Okay, so I have this problem where I need to find the intersection of two lines. The lines are given by the equations ( y = -3x + 1 ) and ( y + 4 = 15x - 2 ). Hmm, I remember that the intersection point is where both equations are true at the same time, meaning the x and y values satisfy both equations. So, I need to solve this system of equations.First, let me write down the two equations clearly:1. ( y = -3x + 1 )2. ( y + 4 = 15x - 2 )I think substitution is a good method here because the first equation already gives me y in terms of x. That should make it easier to substitute into the second equation. So, from the first equation, I can express y as ( y = -3x + 1 ). I can plug this expression for y into the second equation wherever y is.Let me do that. The second equation is ( y + 4 = 15x - 2 ). If I substitute ( y ) with ( -3x + 1 ), it becomes:( (-3x + 1) + 4 = 15x - 2 )Now, I need to simplify this equation step by step. Let me first combine the constants on the left side. So, ( 1 + 4 ) is 5. That gives me:( -3x + 5 = 15x - 2 )Okay, now I need to solve for x. I'll try to get all the x terms on one side and the constants on the other. Let me add ( 3x ) to both sides to eliminate the ( -3x ) on the left. That would give me:( 5 = 18x - 2 )Wait, let me check that again. If I add ( 3x ) to both sides:Left side: ( -3x + 3x + 5 = 5 )Right side: ( 15x + 3x - 2 = 18x - 2 )Yes, that's correct. So now I have:( 5 = 18x - 2 )Next, I need to get rid of the -2 on the right side. I'll add 2 to both sides:Left side: ( 5 + 2 = 7 )Right side: ( 18x - 2 + 2 = 18x )So now, the equation is:( 7 = 18x )To solve for x, I'll divide both sides by 18:( x = frac{7}{18} )Alright, so I found that ( x = frac{7}{18} ). Now, I need to find the corresponding y value. I can use either of the original equations to substitute x back into. Since the first equation is already solved for y, that might be simpler.Using the first equation ( y = -3x + 1 ), I substitute ( x = frac{7}{18} ):( y = -3 times frac{7}{18} + 1 )Let me compute that. First, multiply -3 by ( frac{7}{18} ):( -3 times frac{7}{18} = -frac{21}{18} )Simplify ( -frac{21}{18} ). Both numerator and denominator are divisible by 3:( -frac{21 ÷ 3}{18 ÷ 3} = -frac{7}{6} )Wait, that doesn't seem right. Let me double-check. ( -3 times frac{7}{18} ) is indeed ( -frac{21}{18} ), which simplifies to ( -frac{7}{6} ) because 21 divided by 3 is 7 and 18 divided by 3 is 6. So, that's correct.Now, add 1 to ( -frac{7}{6} ). To add these, I need a common denominator. 1 can be written as ( frac{6}{6} ):( y = -frac{7}{6} + frac{6}{6} = frac{-7 + 6}{6} = frac{-1}{6} )So, ( y = -frac{1}{6} ).Wait a second, let me verify this with the second equation to make sure I didn't make a mistake. The second equation is ( y + 4 = 15x - 2 ). Let me plug in ( x = frac{7}{18} ) and ( y = -frac{1}{6} ) into this equation.Left side: ( y + 4 = -frac{1}{6} + 4 ). Let's compute that. 4 is ( frac{24}{6} ), so:( -frac{1}{6} + frac{24}{6} = frac{23}{6} )Right side: ( 15x - 2 = 15 times frac{7}{18} - 2 ). Let's compute that. First, ( 15 times frac{7}{18} ).Simplify ( 15 times frac{7}{18} ). 15 and 18 have a common factor of 3. So, divide 15 by 3 to get 5, and 18 by 3 to get 6. So, it becomes ( 5 times frac{7}{6} = frac{35}{6} ).Now, subtract 2 from ( frac{35}{6} ). 2 is ( frac{12}{6} ), so:( frac{35}{6} - frac{12}{6} = frac{23}{6} )So, both sides equal ( frac{23}{6} ), which means my solution is correct.Therefore, the intersection point is ( left( frac{7}{18}, -frac{1}{6} right) ).I think that's it. Let me just recap to make sure I didn't skip any steps or make any calculation errors. I substituted y from the first equation into the second, solved for x, then found y using the value of x. Then, I double-checked by plugging both values back into the second equation to ensure consistency. Everything seems to add up, so I'm confident in my answer.