💡First, I need to determine the slope of the line. Since the slope angle is (45^circ), the slope (m) can be calculated using the tangent function:[m = tan(45^circ) = 1]Next, I'll use the point-slope form of a line equation, which is:[y - y_1 = m(x - x_1)]Here, the point (P(-2, 3)) gives (x_1 = -2) and (y_1 = 3). Substituting the known values:[y - 3 = 1(x - (-2)) y - 3 = x + 2 y = x + 5]Finally, I'll compare this equation with the given options to find the correct one.

💡Okay, so I've got this functional equation to solve:[ x f(2 f(y) - x) + y^2 f(2 x - f(y)) = frac{f(x)^2}{x} + f(y f(y)) ]for all integers ( x ) and ( y ) where ( x neq 0 ). Hmm, functional equations can be tricky, but let me try to break this down step by step.First, I remember that a common strategy is to substitute specific values for ( x ) or ( y ) to simplify the equation and maybe find some properties of ( f ). Let me try substituting ( y = 0 ) because that often simplifies things.Substituting ( y = 0 ):[ x f(2 f(0) - x) + 0^2 f(2 x - f(0)) = frac{f(x)^2}{x} + f(0 cdot f(0)) ]Simplifying this, the second term on the left becomes zero, and the right side becomes ( frac{f(x)^2}{x} + f(0) ). So we have:[ x f(2 f(0) - x) = frac{f(x)^2}{x} + f(0) ]Hmm, this looks interesting. Let me denote ( c = f(0) ) to make it simpler:[ x f(2c - x) = frac{f(x)^2}{x} + c ]I wonder if I can find ( c ). Maybe if I choose a specific value for ( x ), I can get more information. Let's try ( x = 1 ):[ 1 cdot f(2c - 1) = frac{f(1)^2}{1} + c ][ f(2c - 1) = f(1)^2 + c ]Not sure if that helps directly. Maybe I should try another substitution. How about ( x = 2c )?Wait, but ( x ) has to be an integer, and ( c ) is ( f(0) ), which is also an integer. So if I set ( x = 2c ):[ 2c cdot f(2c - 2c) = frac{f(2c)^2}{2c} + c ][ 2c cdot f(0) = frac{f(2c)^2}{2c} + c ][ 2c cdot c = frac{f(2c)^2}{2c} + c ][ 2c^2 = frac{f(2c)^2}{2c} + c ]Multiply both sides by ( 2c ):[ 4c^3 = f(2c)^2 + 2c^2 ][ f(2c)^2 = 4c^3 - 2c^2 ][ f(2c)^2 = 2c^2(2c - 1) ]Hmm, since ( f(2c) ) is an integer, the right side must be a perfect square. Let's see, ( 2c^2(2c - 1) ) must be a perfect square. Let me denote ( k = c ), so:[ 2k^2(2k - 1) ]This should be a perfect square. Let me think about possible integer values of ( k ) such that this expression is a perfect square.Let me try ( k = 0 ):[ 2(0)^2(2(0) - 1) = 0 ], which is a perfect square (0). So ( c = 0 ) is a possibility.What about ( k = 1 ):[ 2(1)^2(2(1) - 1) = 2(1)(1) = 2 ], which is not a perfect square.( k = 2 ):[ 2(4)(3) = 24 ], not a perfect square.( k = 3 ):[ 2(9)(5) = 90 ], not a perfect square.( k = -1 ):[ 2(1)(-3) = -6 ], negative, so not a square.So only ( k = 0 ) gives a perfect square. Therefore, ( c = f(0) = 0 ).Alright, so ( f(0) = 0 ). Let me plug this back into the equation I had earlier:[ x f(-x) = frac{f(x)^2}{x} ]Multiply both sides by ( x ):[ x^2 f(-x) = f(x)^2 ]So, ( f(-x) = frac{f(x)^2}{x^2} ). Hmm, interesting. So ( f(-x) ) is related to ( f(x) ) squared divided by ( x^2 ). Since ( f ) maps integers to integers, ( frac{f(x)^2}{x^2} ) must also be an integer. Therefore, ( x^2 ) divides ( f(x)^2 ). Which implies that ( x ) divides ( f(x) ), because if ( x^2 ) divides ( f(x)^2 ), then ( x ) divides ( f(x) ).So, ( f(x) ) is a multiple of ( x ). Let me denote ( f(x) = x cdot g(x) ), where ( g(x) ) is some function from integers to integers.Let me substitute ( f(x) = x g(x) ) into the equation ( x^2 f(-x) = f(x)^2 ):[ x^2 (-x) g(-x) = (x g(x))^2 ][ -x^3 g(-x) = x^2 g(x)^2 ]Divide both sides by ( x^2 ) (since ( x neq 0 )):[ -x g(-x) = g(x)^2 ]So, ( g(-x) = -frac{g(x)^2}{x} )Hmm, so ( g(-x) ) is related to ( g(x) ) squared divided by ( x ). But ( g(-x) ) must be an integer, so ( x ) must divide ( g(x)^2 ). So, ( x ) divides ( g(x)^2 ). Since ( x ) is an integer, ( g(x) ) must be a multiple of ( sqrt{x} ), but since we're dealing with integers, ( x ) must be a square number for ( g(x) ) to be an integer. Wait, but ( x ) can be any integer, positive or negative, so this might complicate things.Alternatively, maybe ( g(x) ) is zero for all ( x ). Let me check that. If ( g(x) = 0 ), then ( f(x) = 0 ) for all ( x ). Let's see if that satisfies the original equation.Substitute ( f(x) = 0 ):Left side:[ x f(2 f(y) - x) + y^2 f(2 x - f(y)) = x f(-x) + y^2 f(2x) = x cdot 0 + y^2 cdot 0 = 0 ]Right side:[ frac{f(x)^2}{x} + f(y f(y)) = frac{0}{x} + f(0) = 0 + 0 = 0 ]So, yes, ( f(x) = 0 ) is a solution.Now, let's see if there are other solutions. Maybe ( f(x) = x^2 ). Let me test that.Substitute ( f(x) = x^2 ):Left side:[ x f(2 f(y) - x) + y^2 f(2 x - f(y)) ][ = x f(2 y^2 - x) + y^2 f(2x - y^2) ][ = x (2 y^2 - x)^2 + y^2 (2x - y^2)^2 ]Right side:[ frac{f(x)^2}{x} + f(y f(y)) ][ = frac{(x^2)^2}{x} + f(y cdot y^2) ][ = frac{x^4}{x} + f(y^3) ][ = x^3 + (y^3)^2 ][ = x^3 + y^6 ]Now, let's compute the left side:First term:[ x (2 y^2 - x)^2 = x (4 y^4 - 4 x y^2 + x^2) = 4 x y^4 - 4 x^2 y^2 + x^3 ]Second term:[ y^2 (2x - y^2)^2 = y^2 (4x^2 - 4 x y^2 + y^4) = 4 x^2 y^2 - 4 x y^4 + y^6 ]Adding both terms:[ 4 x y^4 - 4 x^2 y^2 + x^3 + 4 x^2 y^2 - 4 x y^4 + y^6 ]Simplify:- ( 4 x y^4 - 4 x y^4 = 0 )- ( -4 x^2 y^2 + 4 x^2 y^2 = 0 )- Remaining terms: ( x^3 + y^6 )So, left side equals right side: ( x^3 + y^6 ). Therefore, ( f(x) = x^2 ) is also a solution.Are there any other possible solutions? Let me think. Suppose ( f ) is neither zero nor ( x^2 ). Maybe a combination? For example, ( f(x) = 0 ) for some ( x ) and ( f(x) = x^2 ) for others. But functional equations often require consistency across all inputs, so such piecewise functions might not satisfy the equation for all ( x ) and ( y ).Let me check if ( f(x) = 0 ) and ( f(x) = x^2 ) are the only possible solutions. Suppose there exists some ( x ) where ( f(x) neq 0 ) and ( f(x) neq x^2 ). Then, from earlier, we have ( f(-x) = frac{f(x)^2}{x^2} ). If ( f(x) ) is not zero or ( x^2 ), this might cause issues because ( f(-x) ) must be an integer, so ( x^2 ) must divide ( f(x)^2 ). But if ( f(x) ) is not a multiple of ( x ), this would not hold. Wait, earlier I concluded that ( f(x) ) must be a multiple of ( x ), so ( f(x) = x g(x) ). So, if ( f(x) ) is a multiple of ( x ), then ( g(x) ) must be such that ( g(-x) = -frac{g(x)^2}{x} ). This seems restrictive. Let's suppose ( g(x) = x ). Then ( f(x) = x cdot x = x^2 ), which works. If ( g(x) = 0 ), then ( f(x) = 0 ), which also works. What if ( g(x) ) is something else? Let's say ( g(x) = k x ) for some constant ( k ). Then ( f(x) = k x^2 ). Let me test this.Suppose ( f(x) = k x^2 ). Then:Left side:[ x f(2 f(y) - x) + y^2 f(2x - f(y)) ][ = x f(2 k y^2 - x) + y^2 f(2x - k y^2) ][ = x cdot k (2 k y^2 - x)^2 + y^2 cdot k (2x - k y^2)^2 ][ = k x (4 k^2 y^4 - 4 k x y^2 + x^2) + k y^2 (4x^2 - 4 k x y^2 + k^2 y^4) ][ = 4 k^3 x y^4 - 4 k^2 x^2 y^2 + k x^3 + 4 k x^2 y^2 - 4 k^2 x y^4 + k^3 y^6 ]Simplify:- ( 4 k^3 x y^4 - 4 k^2 x y^4 = 4 k^2 x y^4 (k - 1) )- ( -4 k^2 x^2 y^2 + 4 k x^2 y^2 = 4 k x^2 y^2 (1 - k) )- Remaining terms: ( k x^3 + k^3 y^6 )Right side:[ frac{f(x)^2}{x} + f(y f(y)) ][ = frac{(k x^2)^2}{x} + f(y cdot k y^2) ][ = frac{k^2 x^4}{x} + f(k y^3) ][ = k^2 x^3 + k (k y^3)^2 ][ = k^2 x^3 + k^3 y^6 ]Comparing left and right sides:Left side:[ 4 k^2 x y^4 (k - 1) + 4 k x^2 y^2 (1 - k) + k x^3 + k^3 y^6 ]Right side:[ k^2 x^3 + k^3 y^6 ]For these to be equal for all ( x ) and ( y ), the coefficients of like terms must be equal.Looking at the ( x y^4 ) term:Left: ( 4 k^2 (k - 1) )Right: 0So, ( 4 k^2 (k - 1) = 0 )Which implies ( k = 0 ) or ( k = 1 ).Similarly, looking at the ( x^2 y^2 ) term:Left: ( 4 k (1 - k) )Right: 0So, ( 4 k (1 - k) = 0 )Which also implies ( k = 0 ) or ( k = 1 ).And the ( x^3 ) term:Left: ( k )Right: ( k^2 )So, ( k = k^2 )Which implies ( k = 0 ) or ( k = 1 ).And the ( y^6 ) term:Left: ( k^3 )Right: ( k^3 )Which is consistent for any ( k ).Therefore, the only possible constants ( k ) are 0 and 1, leading to ( f(x) = 0 ) or ( f(x) = x^2 ).So, it seems that the only solutions are ( f(x) = 0 ) and ( f(x) = x^2 ).Let me just verify once more with another substitution. How about ( x = 1 ) and ( y = 1 ):For ( f(x) = 0 ):Left side: ( 1 cdot 0 + 1^2 cdot 0 = 0 )Right side: ( frac{0^2}{1} + f(1 cdot 0) = 0 + 0 = 0 )Okay, works.For ( f(x) = x^2 ):Left side: ( 1 cdot f(2 cdot 1^2 - 1) + 1^2 cdot f(2 cdot 1 - 1^2) )[ = 1 cdot f(2 - 1) + 1 cdot f(2 - 1) ][ = f(1) + f(1) = 1 + 1 = 2 ]Right side: ( frac{1^2}{1} + f(1 cdot 1^2) = 1 + f(1) = 1 + 1 = 2 )Works as well.Another test: ( x = 2 ), ( y = 1 ):For ( f(x) = 0 ):Left: ( 2 cdot 0 + 1^2 cdot 0 = 0 )Right: ( frac{0^2}{2} + f(1 cdot 0) = 0 + 0 = 0 )Good.For ( f(x) = x^2 ):Left: ( 2 f(2 cdot 1^2 - 2) + 1^2 f(2 cdot 2 - 1^2) )[ = 2 f(2 - 2) + 1 f(4 - 1) ][ = 2 f(0) + f(3) = 2 cdot 0 + 9 = 9 ]Right: ( frac{2^2}{2} + f(1 cdot 1^2) = frac{4}{2} + f(1) = 2 + 1 = 3 )Wait, that doesn't match. Hmm, did I make a mistake?Wait, no, let me recalculate:Wait, ( f(2 f(y) - x) ) when ( y = 1 ) and ( x = 2 ):( 2 f(1) - 2 = 2 cdot 1 - 2 = 0 ), so ( f(0) = 0 ).And ( 2x - f(y) = 4 - 1 = 3 ), so ( f(3) = 9 ).So left side: ( 2 cdot 0 + 1 cdot 9 = 9 )Right side: ( frac{4}{2} + f(1) = 2 + 1 = 3 )Wait, that's a problem. 9 ≠ 3. Did I do something wrong?Wait, no, actually, when ( f(x) = x^2 ), substituting into the original equation:Left side: ( x f(2 f(y) - x) + y^2 f(2x - f(y)) )[ = x (2 f(y) - x)^2 + y^2 (2x - f(y))^2 ]Right side: ( frac{f(x)^2}{x} + f(y f(y)) )[ = frac{x^4}{x} + (y f(y))^2 ][ = x^3 + (y cdot y^2)^2 ][ = x^3 + y^6 ]Wait, in the case ( x = 2 ), ( y = 1 ):Left side:[ 2 (2 cdot 1^2 - 2)^2 + 1^2 (2 cdot 2 - 1^2)^2 ][ = 2 (2 - 2)^2 + 1 (4 - 1)^2 ][ = 2 cdot 0 + 1 cdot 9 = 9 ]Right side:[ 2^3 + 1^6 = 8 + 1 = 9 ]Ah, I see, I made a mistake earlier. The right side is ( x^3 + y^6 ), not ( x^3 + (y f(y))^2 ). So, it should be ( 8 + 1 = 9 ), which matches the left side. Phew, that was a close call.So, both solutions ( f(x) = 0 ) and ( f(x) = x^2 ) satisfy the equation.I think I've checked enough cases and substitutions to be confident that these are the only solutions. It seems that the functional equation is satisfied only by the zero function and the square function.

💡First, I need to add the two numbers: 45.768 and 18.1542. To ensure accuracy, I'll align the decimal points and add each corresponding place value.Starting with the thousandths place: 8 plus 2 equals 10. I'll write down 0 and carry over 1 to the hundredths place.Next, in the hundredths place: 6 plus 4 equals 10, plus the carried over 1 makes 11. I'll write down 1 and carry over another 1 to the tenths place.Moving to the tenths place: 7 plus 5 equals 12, plus the carried over 1 makes 13. I'll write down 3 and carry over 1 to the units place.In the units place: 5 plus 8 equals 13, plus the carried over 1 makes 14. I'll write down 4 and carry over 1 to the tens place.Finally, in the tens place: 4 plus 1 equals 5, plus the carried over 1 makes 6.So, the sum of 45.768 and 18.1542 is 63.9222.Now, I need to round this result to the nearest hundredth. The hundredths place is the second digit after the decimal, which is 2. To determine if I round up or down, I'll look at the thousandths place, which is 2. Since 2 is less than 5, I'll keep the hundredths place as it is.Therefore, rounding 63.9222 to the nearest hundredth gives me 63.92.

💡First, I need to solve the equation (frac{a}{2} = frac{3}{b}) to find the value of (ab).To eliminate the fractions, I'll cross-multiply:[a cdot b = 2 cdot 3]This simplifies to:[ab = 6]So, the value of (ab) is 6.

💡First, I recognize that the area of a triangle can be calculated using the formula:[ text{Area} = frac{text{base} times text{height}}{2} ]Given that the area is 615 m² and the height (the perpendicular dropped from the opposite vertex) is 10 meters, I need to find the length of the base.To solve for the base, I rearrange the formula:[ text{base} = frac{2 times text{Area}}{text{height}} ]Substituting the given values:[ text{base} = frac{2 times 615}{10} ][ text{base} = frac{1230}{10} ][ text{base} = 123 ]Therefore, the length of the side is **123 meters**.

💡Okay, so I have this problem to simplify the expression: the fourth root of the cube root of the square root of 1/65536. Hmm, that sounds a bit complicated, but maybe I can break it down step by step.First, let me write down the expression to make it clearer:sqrt[4]{sqrt[3]{sqrt{frac{1}{65536}}}}Alright, so starting from the inside, there's a square root of 1/65536. I remember that square roots can be written as exponents, so the square root of something is the same as raising it to the power of 1/2. Similarly, cube roots are exponents of 1/3, and fourth roots are exponents of 1/4. Maybe if I rewrite everything using exponents, it'll be easier to handle.Let me start with the innermost part: the square root of 1/65536. So that would be:sqrt{frac{1}{65536}} = left(frac{1}{65536}right)^{1/2}Now, I need to figure out what 65536 is in terms of powers of 2 because I remember that 2 raised to some power might give me 65536. Let me think: 2^10 is 1024, 2^16 is 65536. Yeah, that's right. So 65536 is 2^16. Therefore, 1/65536 is 2^-16.So substituting that back in, I have:left(2^{-16}right)^{1/2}When you raise a power to another power, you multiply the exponents. So:2^{-16 times frac{1}{2}} = 2^{-8}Alright, so the square root of 1/65536 is 2^-8, which is 1/256. Okay, moving out a layer, now I have the cube root of that result. So:sqrt[3]{2^{-8}} = left(2^{-8}right)^{1/3}Again, multiplying the exponents:2^{-8 times frac{1}{3}} = 2^{-frac{8}{3}}So the cube root of 2^-8 is 2^(-8/3). Now, moving out to the fourth root, which is the outermost operation:sqrt[4]{2^{-frac{8}{3}}} = left(2^{-frac{8}{3}}right)^{1/4}Multiplying the exponents again:2^{-frac{8}{3} times frac{1}{4}} = 2^{-frac{8}{12}} = 2^{-frac{2}{3}}Simplifying 2^(-2/3) is the same as 1 over 2^(2/3). And 2^(2/3) is the cube root of 2 squared, which is the cube root of 4. So:2^{-frac{2}{3}} = frac{1}{sqrt[3]{4}}Therefore, the entire expression simplifies to 1 over the cube root of 4. I think that's as simplified as it gets because 4 is 2 squared, and the cube root doesn't simplify further into a whole number or a simpler fraction.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the inside:1. Square root of 1/65536: 65536 is 2^16, so square root is 2^(-16 * 1/2) = 2^-8, which is 1/256. That seems right.2. Cube root of 1/256: 256 is 2^8, so cube root is 2^(-8 * 1/3) = 2^(-8/3). That looks correct.3. Fourth root of 2^(-8/3): 2^(-8/3 * 1/4) = 2^(-2/3). Yep, that's correct.4. 2^(-2/3) is 1 over 2^(2/3), which is 1 over the cube root of 4. Perfect.So, I think my steps are all correct. I don't see any errors in my calculations. So the final answer should be 1 over the cube root of 4, which can be written as:frac{1}{sqrt[3]{4}}I think that's the simplest form as a common fraction. There's no need to rationalize the denominator here because the problem just asks for a common fraction, and this is already a fraction with a radical in the denominator, which is acceptable unless specified otherwise.Just to make sure, maybe I can express 4 as 2^2, so:frac{1}{sqrt[3]{2^2}} = 2^{-2/3}But since the question asks for a common fraction, writing it as 1 over the cube root of 4 is probably the most straightforward way. Yeah, I think that's the answer they're looking for.