answer：Okay, so I have this problem here: "Determine the nature of the area of a square when the length of its side is a rational number. The area is: (A) rational (B) irrational (C) integral (D) a perfect square (E) none of these." Alright, let's break this down step by step.First, I know that the area of a square is calculated by squaring the length of its side. So, if the side length is 's', then the area 'A' is s squared, or A = s². Now, the problem states that the side length 's' is a rational number. What's a rational number? A rational number is any number that can be expressed as the quotient of two integers, where the denominator is not zero. So, if 's' is rational, it can be written as s = p/q, where p and q are integers, and q ≠ 0.If I substitute s = p/q into the area formula, I get A = (p/q)² = p²/q². Now, p² and q² are both integers because the square of an integer is still an integer. So, p²/q² is the ratio of two integers, which by definition is a rational number. Wait, but let me make sure I'm not missing anything here. Is there a possibility that p²/q² could be something else, like an integer or an irrational number? Well, if q is 1, then p²/q² simplifies to p², which is an integer. So, in that case, the area would be an integer. But the problem doesn't specify that q is 1; it just says the side length is rational. So, the area could be an integer if the denominator q is 1, but it doesn't have to be. What about being a perfect square? A perfect square is an integer that is the square of another integer. Since p²/q² is a rational number, unless q is 1, it won't be an integer, let alone a perfect square. So, if q is not 1, the area won't be a perfect square. Is there a chance the area could be irrational? Well, irrational numbers cannot be expressed as the ratio of two integers. Since we've established that the area is p²/q², which is a ratio of integers, it can't be irrational. So, the area must be rational. But hold on, let me think about this again. If s is rational, then s² is also rational because the square of a rational number is always rational. So, regardless of the specific values of p and q, as long as s is rational, s² will be rational. Therefore, the area must be rational. It doesn't have to be an integer unless the denominator q is 1, and it doesn't have to be a perfect square unless q is 1 and p is an integer. So, the most accurate answer is that the area is rational.Wait, but let me check if there's any exception or special case where this might not hold. Suppose p and q are such that p²/q² simplifies to an integer or a perfect square. For example, if p = 2 and q = 1, then p²/q² = 4, which is both an integer and a perfect square. But if p = 1 and q = 2, then p²/q² = 1/4, which is rational but neither an integer nor a perfect square. So, depending on the values of p and q, the area can sometimes be an integer or a perfect square, but it's not always the case. The only consistent and guaranteed nature of the area is that it's rational. Therefore, the correct answer should be (A) rational.

answer：First, I need to determine the value of ( p ) in the equation ( y = 2px^2 ) using the fact that the parabola passes through the point ( (1, 4) ).Substituting ( x = 1 ) and ( y = 4 ) into the equation:[4 = 2p(1)^2][4 = 2p][p = 2]Now that I have ( p = 2 ), the equation of the parabola becomes:[y = 4x^2]Next, I'll rewrite this equation in the standard form of a parabola that opens upward:[x^2 = frac{1}{4}y]In the standard form ( x^2 = 4py ), the focus is located at ( (0, p) ). Comparing this with my equation:[4p = frac{1}{4}][p = frac{1}{16}]Therefore, the coordinates of the focus are ( left(0, frac{1}{16}right) ).

answer：Okay, so I have this problem about an equilateral triangle ABC, and O is the center of its inscribed circle. The area of the circle is given as 4π square centimeters, and I need to find the area of the triangle ABC. Hmm, let me think about how to approach this.First, I remember that the area of a circle is given by the formula A = πr², where r is the radius. Since the area is 4π, I can set up the equation πr² = 4π. If I divide both sides by π, that cancels out, and I get r² = 4. Taking the square root of both sides, I find that r = 2 cm. So, the radius of the inscribed circle is 2 cm.Now, since O is the center of the inscribed circle (also called the incenter) of the equilateral triangle ABC, I know that this inradius is related to the sides of the triangle. I recall that in an equilateral triangle, the inradius (r) can be expressed in terms of the side length (a) using the formula r = (a√3)/6. Let me write that down:r = (a√3)/6I already found that r = 2 cm, so plugging that into the equation:2 = (a√3)/6To solve for a, I can multiply both sides by 6:2 * 6 = a√312 = a√3Now, to isolate a, I divide both sides by √3:a = 12 / √3But I remember that it's not good practice to have a square root in the denominator, so I'll rationalize the denominator by multiplying numerator and denominator by √3:a = (12 * √3) / (√3 * √3)a = (12√3) / 3a = 4√3 cmSo, each side of the equilateral triangle is 4√3 cm.Now, I need to find the area of the equilateral triangle. I remember that the area (A) of an equilateral triangle with side length a is given by the formula:A = (√3 / 4) * a²Plugging in the value of a that I found:A = (√3 / 4) * (4√3)²First, let's compute (4√3)²:(4√3)² = 4² * (√3)² = 16 * 3 = 48Now, substitute that back into the area formula:A = (√3 / 4) * 48Simplify this:A = (√3 / 4) * 48 = (√3) * 12 = 12√3So, the area of triangle ABC is 12√3 square centimeters.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the radius, I used the formula for the inradius of an equilateral triangle correctly. Then, I solved for the side length a, which came out to 4√3 cm. Plugging that into the area formula, I squared it correctly to get 48, and then multiplied by √3 / 4 to get 12√3. That seems right.Alternatively, I can think about the relationship between the inradius and the height of the triangle. In an equilateral triangle, the inradius is one-third of the height. So, if the inradius is 2 cm, the height (h) should be 3 times that, which is 6 cm.Then, the area can also be calculated as (base * height) / 2. The base is 4√3 cm, and the height is 6 cm, so:Area = (4√3 * 6) / 2 = (24√3) / 2 = 12√3 cm²That's the same result as before, so it seems consistent. I think I'm confident that 12√3 is the correct area of the triangle.**Final Answer**The area of triangle ABC is boxed{12sqrt{3}} square centimeters.

answer：Okay, so I have this problem: I need to find the remainder when the polynomial ( x^4 + 2x^2 + 5 ) is divided by ( x - 2 ). Hmm, I remember something about the Remainder Theorem from my algebra class. Let me try to recall how that works.Alright, the Remainder Theorem says that if you divide a polynomial ( p(x) ) by ( x - a ), the remainder is just ( p(a) ). So in this case, since I'm dividing by ( x - 2 ), that means ( a ) is 2. So, I should substitute ( x = 2 ) into the polynomial ( x^4 + 2x^2 + 5 ) and calculate the result. That should give me the remainder.Let me write that down step by step to make sure I don't make any mistakes. First, substitute ( x = 2 ) into each term:1. The first term is ( x^4 ). Plugging in 2, that becomes ( 2^4 ). I know that ( 2^4 ) is 16 because ( 2 times 2 times 2 times 2 = 16 ).2. The second term is ( 2x^2 ). Substituting 2 for x, that becomes ( 2 times (2)^2 ). Let me calculate ( (2)^2 ) first, which is 4. Then multiply by 2, so that's ( 2 times 4 = 8 ).3. The third term is just 5, and since there's no x in it, substituting 2 doesn't change anything. So that remains 5.Now, I need to add all these results together to find the remainder. So, adding them up:16 (from the first term) + 8 (from the second term) + 5 (from the third term) equals... let me do the addition step by step to avoid errors.16 + 8 is 24, and then 24 + 5 is 29. So, the total is 29.Wait, let me double-check my calculations to make sure I didn't make any mistakes. It's easy to mix up numbers when adding and multiplying.Starting again:- ( 2^4 = 16 ) – that's correct.- ( 2 times (2^2) = 2 times 4 = 8 ) – that's right.- The constant term is 5 – no issues there.Adding them up: 16 + 8 is definitely 24, and 24 + 5 is 29. Yep, that seems right.Just to be thorough, maybe I can try another method to confirm. I could perform polynomial long division of ( x^4 + 2x^2 + 5 ) by ( x - 2 ) and see if the remainder is indeed 29. But that might take a while, and I'm pretty confident in the Remainder Theorem method. Still, let me outline how polynomial long division would work here to verify.First, set up the division: dividing ( x^4 + 0x^3 + 2x^2 + 0x + 5 ) by ( x - 2 ). I added the missing terms with coefficients of zero to make it easier.1. Divide the leading term ( x^4 ) by ( x ) to get ( x^3 ). Multiply ( x^3 ) by ( x - 2 ) to get ( x^4 - 2x^3 ).2. Subtract this from the original polynomial: ( (x^4 + 0x^3 + 2x^2 + 0x + 5) - (x^4 - 2x^3) = 2x^3 + 2x^2 + 0x + 5 ).3. Now, divide ( 2x^3 ) by ( x ) to get ( 2x^2 ). Multiply ( 2x^2 ) by ( x - 2 ) to get ( 2x^3 - 4x^2 ).4. Subtract this from the current polynomial: ( (2x^3 + 2x^2 + 0x + 5) - (2x^3 - 4x^2) = 6x^2 + 0x + 5 ).5. Divide ( 6x^2 ) by ( x ) to get ( 6x ). Multiply ( 6x ) by ( x - 2 ) to get ( 6x^2 - 12x ).6. Subtract this from the current polynomial: ( (6x^2 + 0x + 5) - (6x^2 - 12x) = 12x + 5 ).7. Divide ( 12x ) by ( x ) to get 12. Multiply 12 by ( x - 2 ) to get ( 12x - 24 ).8. Subtract this from the current polynomial: ( (12x + 5) - (12x - 24) = 29 ).So, after performing the polynomial long division, the remainder is indeed 29. That matches the result I got using the Remainder Theorem. Therefore, I can be confident that the remainder when ( x^4 + 2x^2 + 5 ) is divided by ( x - 2 ) is 29.I think I've covered both methods to confirm the answer. The Remainder Theorem is definitely the quicker way, but it's good to verify with another method to ensure there were no calculation errors. Everything checks out, so I'm pretty sure the remainder is 29.

answer：Okay, so I have this math problem here: If x is 5/7 and y is 7/5, I need to find the value of (1/3)x^8 y^9 plus 1/7. Hmm, let's break this down step by step. First, I need to compute x^8 and y^9 separately. Since x is 5/7, x^8 would be (5/7) raised to the 8th power. Similarly, y is 7/5, so y^9 is (7/5) raised to the 9th power. Wait, that seems a bit complicated. Maybe there's a smarter way to handle the exponents without calculating each power separately. Let me think. If I multiply x^8 and y^9 together, perhaps some terms will cancel out. Let's try that.So, x^8 is (5/7)^8, which is 5^8 over 7^8. And y^9 is (7/5)^9, which is 7^9 over 5^9. If I multiply these two fractions together, I get (5^8 * 7^9) divided by (7^8 * 5^9). Now, looking at the numerator and the denominator, I can see that 5^8 cancels out with part of 5^9 in the denominator, leaving me with 5 in the denominator. Similarly, 7^9 in the numerator cancels out with 7^8 in the denominator, leaving me with 7 in the numerator. So, after simplifying, x^8 y^9 equals 7/5. Okay, that makes things easier. Now, I need to multiply this result by 1/3. So, 1/3 times 7/5 is 7/15. Next, I have to add 1/7 to this result. So, I have 7/15 plus 1/7. To add these two fractions, I need a common denominator. The least common denominator of 15 and 7 is 105. Converting 7/15 to 105ths, I multiply numerator and denominator by 7, which gives me 49/105. Similarly, converting 1/7 to 105ths, I multiply numerator and denominator by 15, resulting in 15/105. Now, adding these together: 49/105 plus 15/105 equals 64/105. So, putting it all together, the value of (1/3)x^8 y^9 plus 1/7 is 64/105. Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with x^8 y^9, I correctly simplified it to 7/5. Then, multiplying by 1/3 gives 7/15. Adding 1/7, I found the common denominator correctly as 105, and converted both fractions properly. Adding them together gives 64/105, which seems right. I think that's the correct answer. It all checks out.

answer：Alright, so I have this problem here about a parabola and a line. Let me try to figure it out step by step. First, the equation of the parabola is given as ( y^2 = ax ), where ( a neq 0 ). I remember that for a parabola in the form ( y^2 = 4px ), the focus is at ( (p, 0) ). So, comparing that to our equation ( y^2 = ax ), it seems like ( 4p = a ), which means ( p = frac{a}{4} ). Therefore, the focus ( F ) of this parabola should be at ( left( frac{a}{4}, 0 right) ).Next, the problem says there's a line with a slope of 2 that passes through this focus ( F ). So, the equation of this line can be written using the point-slope form. Since it has a slope of 2 and passes through ( F ), the equation should be:( y - 0 = 2 left( x - frac{a}{4} right) )Simplifying that, we get:( y = 2x - frac{a}{2} )This line intersects the y-axis at point ( A ). To find the coordinates of ( A ), I know that any point on the y-axis has an x-coordinate of 0. So, substituting ( x = 0 ) into the equation of the line:( y = 2(0) - frac{a}{2} = -frac{a}{2} )Therefore, point ( A ) is ( left( 0, -frac{a}{2} right) ).Now, we need to find the area of triangle ( OAF ), where ( O ) is the origin. The coordinates of the three points are:- ( O = (0, 0) )- ( A = left( 0, -frac{a}{2} right) )- ( F = left( frac{a}{4}, 0 right) )I remember that the area of a triangle given three points can be calculated using the formula:( text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| )Plugging in the coordinates:( text{Area} = frac{1}{2} |0 left( -frac{a}{2} - 0 right) + 0 left( 0 - 0 right) + frac{a}{4} left( 0 - left( -frac{a}{2} right) right)| )Simplifying each term:- The first term is 0 because it's multiplied by 0.- The second term is also 0 for the same reason.- The third term is ( frac{a}{4} times frac{a}{2} = frac{a^2}{8} )So, the area becomes:( frac{1}{2} times left| frac{a^2}{8} right| = frac{a^2}{16} )But the problem states that the area is 4. Therefore:( frac{a^2}{16} = 4 )Multiplying both sides by 16:( a^2 = 64 )Taking the square root of both sides:( a = pm 8 )So, the equation of the parabola is ( y^2 = ax ), which becomes ( y^2 = pm 8x ). Looking at the options, that's option B.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the focus correctly as ( left( frac{a}{4}, 0 right) ).2. Wrote the equation of the line with slope 2 through ( F ): ( y = 2x - frac{a}{2} ). That seems right.3. Found point ( A ) by setting ( x = 0 ): ( y = -frac{a}{2} ). Makes sense.4. Calculated the area using the formula. Let me verify that formula again. Alternatively, since two points are on the axes, maybe I can use a simpler method.Another way to calculate the area is to note that triangle ( OAF ) is a right triangle with base along the x-axis from ( O ) to ( F ), which is ( frac{a}{4} ), and height along the y-axis from ( O ) to ( A ), which is ( frac{a}{2} ) (since distance can't be negative). So, area is ( frac{1}{2} times text{base} times text{height} ):( text{Area} = frac{1}{2} times frac{a}{4} times frac{a}{2} = frac{a^2}{16} )Yes, same result. So, setting that equal to 4:( frac{a^2}{16} = 4 ) leads to ( a^2 = 64 ), so ( a = pm 8 ). Therefore, the equation is ( y^2 = pm 8x ), which is option B. I think that's correct.**Final Answer**boxed{B}