💡First, I need to calculate (4^3) and (8^2) separately.Calculating (4^3):[4^3 = 4 times 4 times 4 = 64]Calculating (8^2):[8^2 = 8 times 8 = 64]Now, subtract the second result from the first:[4^3 - 8^2 = 64 - 64 = 0]Therefore, the final answer is (0).

💡First, I recognize that the expression (dfrac{10!}{7!3!}) involves factorials, which are products of consecutive integers up to a certain number.I can simplify this expression by canceling out the common factorial terms in the numerator and the denominator. Specifically, the (7!) in the numerator and the (7!) in the denominator will cancel each other out.After canceling, the expression simplifies to (dfrac{10 times 9 times 8}{3 times 2 times 1}).Next, I calculate the numerator: (10 times 9 times 8 = 720).Then, I calculate the denominator: (3 times 2 times 1 = 6).Finally, I divide the numerator by the denominator: (dfrac{720}{6} = 120).Therefore, the value of the expression is 120.

💡First, I need to determine the car's fuel efficiency in the city, denoted as ( C ) miles per gallon (mpg). I know that the car's fuel efficiency on the highway is 12 mpg higher than in the city, so I can express the highway fuel efficiency as ( H = C + 12 ).Next, I'll use the information about the miles per tankful to set up equations. On the highway, the car travels 800 miles per tank, and in the city, it travels 500 miles per tank. Let ( T ) represent the size of the gas tank in gallons.For the highway:[800 = H times T]Substituting ( H = C + 12 ):[800 = (C + 12) times T]For the city:[500 = C times T]Now, I'll solve both equations for ( T ):[T = frac{800}{C + 12}][T = frac{500}{C}]Since both expressions equal ( T ), I can set them equal to each other:[frac{800}{C + 12} = frac{500}{C}]Cross-multiplying to solve for ( C ):[800C = 500(C + 12)][800C = 500C + 6000][300C = 6000][C = 20]Therefore, the car travels 20 miles per gallon in the city.

💡Alright, so I have this function f(x) = x|x| + bx + c, and there are four propositions about it. I need to figure out which ones are correct. Let me go through each proposition one by one.Starting with proposition ①: When C = 0, y = f(x) is an odd function. Hmm, okay. If C is 0, then the function simplifies to f(x) = x|x| + bx. I remember that an odd function satisfies f(-x) = -f(x). Let me test that. If I plug in -x into f(x), I get f(-x) = (-x)|-x| + b(-x) = -x|x| - bx. That's exactly -f(x) because f(x) = x|x| + bx, so -f(x) = -x|x| - bx. Yep, that checks out. So proposition ① is correct.Moving on to proposition ②: When b = 0 and c > 0, the equation f(x) = 0 has only one real root. Okay, so if b is 0, the function becomes f(x) = x|x| + c. Setting that equal to zero: x|x| + c = 0. Since c is positive, we have x|x| = -c. But x|x| is always non-negative because |x| is non-negative, and multiplying by x, whether positive or negative, gives a non-negative result. So x|x| can never be negative, which means x|x| = -c has no real solutions. Wait, that means there are no real roots, not just one. So proposition ② says there's only one real root, but actually, there are none. So proposition ② is incorrect.Next, proposition ③: The graph of y = f(x) is symmetric about the point (0, c). Symmetry about a point usually means that if you rotate the graph 180 degrees around that point, it looks the same. For a function to be symmetric about (0, c), it should satisfy f(-x) = 2c - f(x). Let me check that. f(-x) = (-x)|-x| + b(-x) + c = -x|x| - bx + c. On the other hand, 2c - f(x) = 2c - (x|x| + bx + c) = 2c - x|x| - bx - c = c - x|x| - bx. Comparing f(-x) and 2c - f(x): f(-x) = -x|x| - bx + c, and 2c - f(x) = c - x|x| - bx. They look the same, so f(-x) = 2c - f(x). That means the graph is symmetric about the point (0, c). So proposition ③ is correct.Finally, proposition ④: The equation f(x) = 0 has at most two real roots. Let's see. The function f(x) = x|x| + bx + c can be rewritten as f(x) = x^2 + bx + c for x ≥ 0 and f(x) = -x^2 + bx + c for x < 0. So for x ≥ 0, it's a quadratic equation, and for x < 0, it's another quadratic equation. Each quadratic can have at most two real roots, but since they're defined on different intervals, the total number of real roots can be up to two from each interval, but actually, since x ≥ 0 and x < 0 are separate, the maximum number of real roots is two. Wait, no, each quadratic can have up to two roots, but depending on the discriminant, they might have one or two. So overall, the equation f(x) = 0 can have up to two real roots. So proposition ④ is correct.Wait, but earlier I thought proposition ② was incorrect because there are no real roots when b = 0 and c > 0. But the proposition says it has only one real root. So maybe I made a mistake there. Let me double-check. If b = 0 and c > 0, f(x) = x|x| + c. Setting that to zero: x|x| = -c. Since x|x| is always non-negative, and c is positive, x|x| = -c has no solution. So there are no real roots, not one. Therefore, proposition ② is incorrect.But in the standard solution, it says propositions ①, ②, and ③ are correct. Maybe I'm misunderstanding something. Let me think again about proposition ②. If b = 0 and c > 0, f(x) = x|x| + c. Maybe there's a root at x = 0? Plugging in x = 0: f(0) = 0 + 0 + c = c ≠ 0. So no, x = 0 is not a root. So there are no real roots. Therefore, proposition ② is incorrect.Hmm, maybe the standard solution is wrong? Or maybe I'm missing something. Let me check proposition ③ again. The graph is symmetric about (0, c). I thought f(-x) = 2c - f(x), which is the condition for symmetry about (0, c). Let me verify with specific values. Suppose c = 1, b = 0. Then f(x) = x|x| + 1. So f(1) = 1*1 + 1 = 2, f(-1) = -1*1 + 1 = 0. Now, 2c - f(1) = 2*1 - 2 = 0, which equals f(-1). Similarly, f(2) = 4 + 1 = 5, f(-2) = -4 + 1 = -3. 2c - f(2) = 2 - 5 = -3, which equals f(-2). So it seems to hold. Therefore, proposition ③ is correct.So, in conclusion, propositions ①, ③, and ④ are correct, while proposition ② is incorrect. But the standard solution says ①, ②, and ③ are correct. Maybe there's a different interpretation of proposition ②. Perhaps when b = 0 and c > 0, the equation f(x) = 0 has only one real root at x = 0, but f(0) = c ≠ 0. So that doesn't make sense. Maybe I'm missing something else.Wait, maybe when b = 0 and c > 0, the equation f(x) = 0 has only one real root at x = -c? Let's see: f(-c) = (-c)|-c| + 0 + c = -c^2 + c. Setting that equal to zero: -c^2 + c = 0 ⇒ c(-c + 1) = 0 ⇒ c = 0 or c = 1. But c > 0, so c = 1. So if c = 1, f(-1) = -1 + 1 = 0. So when c = 1, x = -1 is a root. But for other values of c > 0, like c = 2, f(-2) = -4 + 2 = -2 ≠ 0. So only when c = 1, there's a root at x = -1. Otherwise, no roots. So proposition ② says "when b = 0 and c > 0, the equation f(x) = 0 has only one real root." But this is only true when c = 1, not for all c > 0. Therefore, proposition ② is incorrect in general.So, I think the standard solution might have an error, and the correct propositions are ①, ③, and ④.

💡First, I need to determine how many quarters Sara had initially before her dad gave her additional quarters.I know that after receiving 49 quarters from her dad, Sara has a total of 70 quarters.To find the initial number of quarters, I can subtract the number of quarters her dad gave her from the total number of quarters she has now.So, I subtract 49 quarters from 70 quarters.This calculation will give me the number of quarters Sara had initially.

💡Okay, so I have this problem where a line intersects a circle at two points, M and N. These points are symmetric with respect to the line y = x. I need to find the value of 'a' in the equation of the line y = ax + 1. First, let me make sure I understand the problem correctly. We have a line and a circle, and they intersect at points M and N. These points are symmetric over the line y = x, which means if I reflect one point over y = x, I get the other point. So, if M is (h, k), then N should be (k, h). Now, the circle's equation is given as x² + y² + ax + by - 4 = 0. I remember that the general equation of a circle is x² + y² + Dx + Ey + F = 0, and its center is at (-D/2, -E/2). So, for this circle, the center should be at (-a/2, -b/2). The line y = ax + 1 intersects the circle at M and N. Since M and N are symmetric over y = x, the line y = x should be the perpendicular bisector of the segment MN. That means the center of the circle should lie on the line y = x because the perpendicular bisector of a chord passes through the center of the circle. So, the center of the circle (-a/2, -b/2) must lie on y = x. Therefore, -a/2 = -b/2, which simplifies to a = b. That's one equation relating a and b. Now, I also know that the line y = ax + 1 is perpendicular to the line y = x because the points M and N are symmetric over y = x. Wait, is that necessarily true? Let me think. If two points are symmetric over y = x, the line connecting them is perpendicular to y = x. So, the line MN is perpendicular to y = x, which has a slope of 1. Therefore, the slope of MN should be -1. But the line y = ax + 1 is the line MN. So, the slope 'a' should be -1. Is that correct? Wait, no. Because the line MN is the line y = ax + 1, which is given. So, if the line MN is perpendicular to y = x, then the slope of MN should be -1. Therefore, a = -1. Hmm, that seems straightforward, but let me verify this step by step to make sure I'm not making a mistake. First, the line y = ax + 1 intersects the circle at M and N. Since M and N are symmetric over y = x, the midpoint of MN lies on y = x. Also, the line MN is perpendicular to y = x. So, the slope of y = x is 1, so the slope of MN should be -1 because they are perpendicular. Therefore, the slope of the line y = ax + 1 is a, which must be equal to -1. So, a = -1. But wait, let me think again. The line MN is y = ax + 1, and it's perpendicular to y = x, so yes, their slopes multiply to -1. So, 1 * a = -1, hence a = -1. That seems correct. But let me also check if the center lies on y = x. The center is (-a/2, -b/2). Since a = b, as we found earlier, then the center is (-a/2, -a/2), which is on the line y = x because both coordinates are equal. So, that checks out. Is there anything else I need to consider? Maybe I should substitute a = -1 into the circle equation and see if the points M and N are indeed symmetric over y = x. Let me try that. If a = -1, then the line is y = -x + 1. The circle equation becomes x² + y² - x + by - 4 = 0. Since a = b, b is also -1, so the circle equation is x² + y² - x - y - 4 = 0. Let me find the points of intersection between y = -x + 1 and the circle. Substitute y = -x + 1 into the circle equation:x² + (-x + 1)² - x - (-x + 1) - 4 = 0Let me compute each term:x² + (x² - 2x + 1) - x - (-x + 1) - 4 = 0Simplify term by term:x² + x² - 2x + 1 - x + x - 1 - 4 = 0Combine like terms:2x² - 2x + 1 - x + x - 1 - 4 = 0Simplify:2x² - 2x + 0 - 4 = 0So, 2x² - 2x - 4 = 0Divide both sides by 2:x² - x - 2 = 0Factor:(x - 2)(x + 1) = 0So, x = 2 or x = -1Now, find y for each x:If x = 2, y = -2 + 1 = -1, so point is (2, -1)If x = -1, y = 1 + 1 = 2, so point is (-1, 2)Now, check if these points are symmetric over y = x. The reflection of (2, -1) over y = x is (-1, 2), which is the other point. So, yes, they are symmetric. Therefore, a = -1 is indeed the correct answer. I think I've covered all the steps and verified the result. It makes sense because the line has to be perpendicular to y = x, and the center has to lie on y = x, which both lead to a = -1.