💡First, I need to calculate the product of the slope ( m ) and the y-intercept ( b ). Given that ( m = frac{3}{4} ) and ( b = -frac{2}{3} ), I can multiply these two values together.Multiplying ( frac{3}{4} ) by ( -frac{2}{3} ) gives:[frac{3}{4} times -frac{2}{3} = -frac{6}{12} = -0.5]Next, I need to determine which range the product ( mb ) falls into. The calculated value of ( mb ) is ( -0.5 ), which is between ( -1 ) and ( 0 ).Therefore, the correct range for ( mb ) is ( -1 < mb < 0 ).

💡Okay, so I have this equation here: (a + b)/(b + c) = (c + d)/(d + a). Hmm, I need to figure out what this implies about a, b, c, and d. The options are about whether a must equal c, or if the sum a + b + c + d must be zero, or if either of those is true, or something else. Let me try to work through this step by step.First, I think I should cross-multiply to eliminate the fractions. So, if I do that, I get:(a + b)(d + a) = (b + c)(c + d)Alright, now I'll expand both sides to see what that gives me.On the left side: (a + b)(d + a) = a*d + a*a + b*d + b*a = a² + ad + ab + bdOn the right side: (b + c)(c + d) = b*c + b*d + c*c + c*d = bc + bd + c² + cdSo now I have:a² + ad + ab + bd = bc + bd + c² + cdHmm, I notice that both sides have a bd term. Maybe I can subtract bd from both sides to simplify:a² + ad + ab = bc + c² + cdOkay, now let me rearrange the terms to bring everything to one side:a² + ad + ab - bc - c² - cd = 0Hmm, this looks a bit messy. Maybe I can factor some terms here. Let me see.Looking at the terms with a: a² + ad + ab. I can factor out an a from these:a(a + d + b)Similarly, looking at the terms with c: -bc - c² - cd. I can factor out a -c from these:-c(b + c + d)So now the equation becomes:a(a + b + d) - c(b + c + d) = 0Hmm, interesting. So, a(a + b + d) = c(b + c + d)Wait, maybe I can factor this differently. Let me see if I can write this as (a - c)(something) = 0.Let me try adding and subtracting ac on both sides to make it factorable.Starting from:a(a + b + d) - c(b + c + d) = 0Add and subtract ac:a(a + b + d) + ac - ac - c(b + c + d) = 0Group the terms:a(a + b + d + c) - c(a + b + d + c) = 0Oh! Now I see that both terms have (a + b + c + d) as a common factor. So, factoring that out:(a - c)(a + b + c + d) = 0Wow, okay, so this product equals zero. That means either (a - c) = 0 or (a + b + c + d) = 0, or both.So, if (a - c) = 0, then a = c.If (a + b + c + d) = 0, then the sum of all four variables is zero.Therefore, either a = c or a + b + c + d = 0, or both.Looking back at the options, that's option C: either a = c or a + b + c + d = 0, or both.Let me double-check to make sure I didn't make any mistakes in my algebra.Starting from the cross-multiplied equation:(a + b)(d + a) = (b + c)(c + d)Expanded correctly to:a² + ad + ab + bd = bc + bd + c² + cdSubtracting bd from both sides:a² + ad + ab = bc + c² + cdRearranged to:a² + ad + ab - bc - c² - cd = 0Factored as:a(a + b + d) - c(b + c + d) = 0Then, by adding and subtracting ac, factored further into:(a - c)(a + b + c + d) = 0Yes, that seems correct. So, my conclusion is that either a = c or the sum a + b + c + d is zero, or both. So, option C is the right answer.**Final Answer**boxed{C}

💡First, I need to determine the initial distance ( x ) that Alex and Max ran while being even with each other.1. **Alex gets ahead by 300 feet:** - Both have covered ( x + 300 ) feet.2. **Max gets ahead by 170 feet:** - The distance covered by both becomes ( x + 300 - 170 = x + 130 ) feet.3. **Alex gets ahead by 440 feet:** - The distance covered by both is now ( x + 130 + 440 = x + 570 ) feet.4. **Max has 3,890 feet left to catch up to Alex:** - Alex has covered ( 5000 - 3890 = 1110 ) feet.5. **Set up the equation:** [ x + 570 = 1110 ]6. **Solve for ( x ):** [ x = 1110 - 570 = 540 ]Therefore, Alex and Max were even with each other for **540 feet** at the beginning of the race.

💡Okay, so I have this equation here: 9 raised to the power of y equals 3 raised to the power of x. And I'm told that y is 6. I need to find out what x is. Alright, let's break this down step by step.First, I know that 9 is a number that can be expressed as a power of 3. Specifically, 9 is 3 squared, right? So, 9 is 3^2. That seems useful because both sides of the equation involve powers of 3 now.So, if I rewrite 9^y as (3^2)^y, the equation becomes (3^2)^y = 3^x. Now, there's a rule of exponents that says when you have a power raised to another power, you multiply the exponents. So, (3^2)^y becomes 3^(2y). That simplifies the left side of the equation to 3^(2y).Now, the equation is 3^(2y) = 3^x. Since both sides have the same base, which is 3, I can set the exponents equal to each other. That means 2y = x.But wait, I know that y is 6. So, plugging that into the equation, I get 2*6 = x. That simplifies to 12 = x.Hmm, let me double-check that to make sure I didn't make any mistakes. Starting from the beginning: 9^y = 3^x. Since 9 is 3^2, substituting that in gives (3^2)^y = 3^x. Applying the power rule, that becomes 3^(2y) = 3^x. Setting the exponents equal, 2y = x. With y = 6, that's 2*6 = 12. So, x should be 12.Just to be thorough, maybe I can plug the numbers back into the original equation to verify. If y is 6 and x is 12, then 9^6 should equal 3^12. Let's see:Calculating 9^6: 9*9 = 81, 81*9 = 729, 729*9 = 6561, 6561*9 = 59049, 59049*9 = 531441, and 531441*9 = 4782969.Now, calculating 3^12: 3*3 = 9, 9*3 = 27, 27*3 = 81, 81*3 = 243, 243*3 = 729, 729*3 = 2187, 2187*3 = 6561, 6561*3 = 19683, 19683*3 = 59049, 59049*3 = 177147, 177147*3 = 531441, and 531441*3 = 1594323.Wait a minute, that doesn't match. 9^6 is 531441, and 3^12 is 531441 as well. Oh, I think I made a mistake in my earlier calculation when I thought 3^12 was 1594323. Let me recount:3^1 = 33^2 = 93^3 = 273^4 = 813^5 = 2433^6 = 7293^7 = 21873^8 = 65613^9 = 196833^10 = 590493^11 = 1771473^12 = 531441Yes, that's correct. So, 3^12 is indeed 531441, which matches 9^6. Therefore, my initial solution was correct: x equals 12.I think I confused myself earlier by miscounting the multiplications, but upon recounting, everything checks out. So, x is definitely 12.

💡First, I will assume the man's original wages are 100 for simplicity.Next, I will calculate the decrease of 50% on the original wages:50% of 100 is 50.Subtracting this decrease from the original wages gives:100 - 50 = 50.Then, I will calculate the increase of 50% on the reduced wages:50% of 50 is 25.Adding this increase to the reduced wages results in:50 + 25 = 75.Finally, to determine the loss, I will subtract the final wages from the original wages:100 - 75 = 25.Therefore, the man has a loss of 25.

💡Okay, so I have this problem here about a fifth-degree polynomial with real coefficients. All the roots are complex numbers lying on a circle centered at the origin with radius 1. I need to find the sum of the reciprocals of the roots, and it's supposed to be one of the coefficients or something related. Let me think through this step by step.First, let's recall that for a polynomial equation, the sum and products of the roots can be related to the coefficients using Vieta's formulas. For a general polynomial ( z^5 + pz^4 + qz^3 + rz^2 + sz + t = 0 ), the sum of the roots is (-p), the sum of the products of the roots taken two at a time is (q), and so on. But here, I need the sum of the reciprocals of the roots, not the roots themselves.Hmm, reciprocals. So if the roots are ( z_1, z_2, z_3, z_4, z_5 ), then the reciprocals are ( 1/z_1, 1/z_2, 1/z_3, 1/z_4, 1/z_5 ). I need to find ( 1/z_1 + 1/z_2 + 1/z_3 + 1/z_4 + 1/z_5 ).Wait, since all the roots lie on the unit circle, meaning their magnitude is 1, right? So ( |z_i| = 1 ) for each root ( z_i ). That means that the reciprocal of each root is just its complex conjugate. Because if ( z = a + bi ), then ( 1/z = overline{z}/|z|^2 ), and since ( |z| = 1 ), ( 1/z = overline{z} ). So, ( 1/z_i = overline{z_i} ).Therefore, the sum of the reciprocals is the same as the sum of the complex conjugates of the roots. So, ( 1/z_1 + 1/z_2 + 1/z_3 + 1/z_4 + 1/z_5 = overline{z_1} + overline{z_2} + overline{z_3} + overline{z_4} + overline{z_5} ).Now, since the polynomial has real coefficients, the complex roots must come in conjugate pairs. That means if ( z ) is a root, then ( overline{z} ) is also a root. So, the sum of the roots ( z_1 + z_2 + z_3 + z_4 + z_5 ) is equal to the sum of their conjugates ( overline{z_1} + overline{z_2} + overline{z_3} + overline{z_4} + overline{z_5} ). But wait, actually, no. The sum of the roots is ( -p ), and the sum of the conjugates is ( -p ) as well because the coefficients are real. So, does that mean the sum of the reciprocals is equal to the sum of the roots?Wait, no. Let me think again. The sum of the roots is ( z_1 + z_2 + z_3 + z_4 + z_5 = -p ). The sum of the reciprocals is ( overline{z_1} + overline{z_2} + overline{z_3} + overline{z_4} + overline{z_5} ). But since the coefficients are real, the sum of the roots is equal to the sum of their conjugates. So, ( overline{z_1} + overline{z_2} + overline{z_3} + overline{z_4} + overline{z_5} = overline{z_1 + z_2 + z_3 + z_4 + z_5} = overline{-p} ). But since ( p ) is real, ( overline{-p} = -p ). Therefore, the sum of the reciprocals is ( -p ).Wait, so that would mean the sum of the reciprocals is ( -p ), which is option E. Let me double-check this reasoning.Given that all roots are on the unit circle, their reciprocals are their conjugates. The polynomial has real coefficients, so the sum of the roots is ( -p ), and the sum of the conjugates is also ( -p ). Therefore, the sum of the reciprocals is ( -p ). Yeah, that seems correct.Alternatively, I could think about the reciprocal polynomial. If I reverse the coefficients, I get ( t z^5 + s z^4 + r z^3 + q z^2 + p z + 1 = 0 ). But since all roots are on the unit circle, the reciprocal polynomial should have the same roots as the original polynomial. But I'm not sure if that helps directly here.Wait, but if I consider the sum of the reciprocals, that's equivalent to the sum of the roots of the reciprocal polynomial. But the reciprocal polynomial has coefficients ( t, s, r, q, p, 1 ). So, the sum of the roots of the reciprocal polynomial is ( -s/t ). But since all roots are on the unit circle, ( t = 1 ) because the product of the roots is ( t ) (up to a sign), and since each root has magnitude 1, the product is 1. So, ( t = 1 ). Therefore, the sum of the reciprocals is ( -s ). Wait, that contradicts my earlier conclusion.Hmm, maybe I made a mistake here. Let me see. The reciprocal polynomial is ( z^5 + (s/t) z^4 + (r/t) z^3 + (q/t) z^2 + (p/t) z + 1/t = 0 ). If all roots of the original polynomial are on the unit circle, then the reciprocal polynomial has the same roots as the original polynomial. Therefore, the sum of the roots of the reciprocal polynomial is ( -s/t ). But since the roots are the same, the sum of the roots of the reciprocal polynomial should be equal to the sum of the roots of the original polynomial, which is ( -p ). Therefore, ( -s/t = -p ), so ( s/t = p ). But since ( t = 1 ) (because the product of the roots is ( t ) and each root has magnitude 1, so the product is 1), then ( s = p ). Therefore, the sum of the reciprocals is ( -s = -p ). So, that still gives me ( -p ). So, both methods lead to the same conclusion.Wait, so in the reciprocal polynomial, the sum of the roots is ( -s/t ), which is ( -s ) since ( t = 1 ). But the sum of the roots of the reciprocal polynomial is the same as the sum of the reciprocals of the original polynomial's roots. Therefore, ( -s = ) sum of reciprocals. But earlier, I thought it was ( -p ). So, which one is correct?Wait, maybe I confused something. Let's clarify. The sum of the reciprocals of the roots of the original polynomial is equal to the sum of the roots of the reciprocal polynomial. The reciprocal polynomial is ( t z^5 + s z^4 + r z^3 + q z^2 + p z + 1 = 0 ). So, the sum of the roots of the reciprocal polynomial is ( -s/t ). Since the original polynomial has all roots on the unit circle, the reciprocal polynomial has the same roots, so the sum of the roots of the reciprocal polynomial is equal to the sum of the roots of the original polynomial, which is ( -p ). Therefore, ( -s/t = -p ). Since ( t = 1 ), this implies ( s = p ). Therefore, the sum of the reciprocals is ( -s = -p ). So, both ways, it's ( -p ).Alternatively, going back to the original approach, since each reciprocal is the conjugate, and the sum of the conjugates is the conjugate of the sum, which is ( -p ) because ( p ) is real. So, that also gives ( -p ).Therefore, I think the answer is ( -p ), which is option E.Wait, just to make sure, let me think of a simple example. Suppose I have a polynomial with roots on the unit circle, say ( z^2 - (z + 1/z) z + 1 = 0 ). Wait, that's a quadratic. Let me take a specific case. Let’s say the polynomial is ( z^2 - (z + 1/z) z + 1 = 0 ). Wait, that simplifies to ( z^2 - (z + 1/z) z + 1 = z^2 - (z^2 + 1) + 1 = z^2 - z^2 - 1 + 1 = 0 ). Hmm, that's trivial. Maybe a better example.Consider ( z^2 - (z + 1/z) z + 1 = 0 ). Wait, that's the same as ( z^2 - z^2 - 1 + 1 = 0 ). Hmm, not helpful. Maybe take a specific root, say ( z = e^{itheta} ). Then ( 1/z = e^{-itheta} ). So, the sum of reciprocals is ( e^{-itheta} + e^{-iphi} ) if the roots are ( e^{itheta} ) and ( e^{iphi} ). The sum of the roots is ( e^{itheta} + e^{iphi} ), and the sum of the reciprocals is ( e^{-itheta} + e^{-iphi} ), which is the conjugate of the sum of the roots. Since the sum of the roots is ( -p ), the sum of the reciprocals is ( -p ) because ( p ) is real.Yes, that makes sense. So, in the quadratic case, the sum of the reciprocals is the conjugate of the sum of the roots, which is equal to the sum of the roots because the sum is real. Wait, no, in the quadratic case, the sum of the roots is ( -p ), which is real, so the sum of the reciprocals is also ( -p ). So, it works.Therefore, in the fifth-degree case, the same logic applies. The sum of the reciprocals is the sum of the conjugates, which is the conjugate of the sum of the roots. Since the sum of the roots is real (because the coefficients are real), the conjugate is the same as the sum. Therefore, the sum of the reciprocals is equal to the sum of the roots, which is ( -p ). Wait, no, hold on. The sum of the roots is ( -p ), and the sum of the reciprocals is the sum of the conjugates, which is ( overline{z_1 + z_2 + z_3 + z_4 + z_5} = overline{-p} = -p ) because ( p ) is real. Therefore, the sum of the reciprocals is ( -p ).Yes, that seems consistent. So, the answer should be ( -p ), which is option E.