💡Okay, so I need to find the domain of this rational function f(x) = (x³ - 3x² + 5x - 2)/(x³ - 5x² + 8x - 4). Hmm, I remember that for rational functions, the domain is all real numbers except where the denominator is zero because you can't divide by zero. So, my main task here is to figure out where the denominator equals zero and then exclude those x-values from the domain.Alright, let me write down the denominator separately: q(x) = x³ - 5x² + 8x - 4. I need to solve q(x) = 0, which means solving x³ - 5x² + 8x - 4 = 0. Factoring this cubic equation might be a bit tricky, but maybe I can try to find rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, the constant term is -4, and the leading coefficient is 1, so possible roots are ±1, ±2, ±4.Let me test x=1 first. Plugging into q(1): 1 - 5 + 8 - 4 = 0. Oh, that works! So, (x - 1) is a factor. Now, I can perform polynomial division or use synthetic division to factor out (x - 1) from q(x).Using synthetic division with x=1:1 | 1 -5 8 -4 1 -4 4 1 -4 4 0So, after factoring out (x - 1), we have q(x) = (x - 1)(x² - 4x + 4). Now, let's factor the quadratic: x² - 4x + 4. That looks like a perfect square, right? It factors into (x - 2)². So, putting it all together, q(x) = (x - 1)(x - 2)².Wait, so the denominator factors into (x - 1)(x - 2)². That means the denominator is zero when x = 1 or x = 2. But since (x - 2) is squared, does that affect anything? Hmm, I think for the domain, it's still just the values where the denominator is zero, regardless of multiplicity. So, x cannot be 1 or 2.Wait, but hold on. Let me double-check my factoring. I had x³ - 5x² + 8x - 4. Factored as (x - 1)(x² - 4x + 4), which is (x - 1)(x - 2)². Let me expand that to make sure: (x - 1)(x - 2)² = (x - 1)(x² - 4x + 4) = x³ - 4x² + 4x - x² + 4x - 4 = x³ - 5x² + 8x - 4. Yep, that matches. So, my factoring is correct.Therefore, the denominator is zero when x = 1 or x = 2. So, these are the points where the function f(x) is undefined. Therefore, the domain of f(x) is all real numbers except x = 1 and x = 2.But wait, let me check if the numerator also has these roots. The numerator is x³ - 3x² + 5x - 2. Let me see if x=1 is a root: 1 - 3 + 5 - 2 = 1. So, 1 - 3 is -2, plus 5 is 3, minus 2 is 1. So, x=1 is not a root of the numerator. What about x=2: 8 - 12 + 10 - 2 = 4. So, 8 -12 is -4, plus 10 is 6, minus 2 is 4. So, x=2 is not a root of the numerator either. Therefore, both x=1 and x=2 are vertical asymptotes or points where the function is undefined, not holes.So, the domain is all real numbers except x=1 and x=2. In interval notation, that would be (-∞, 1) union (1, 2) union (2, ∞).Wait, but in the initial problem statement, the denominator was x³ - 5x² + 8x - 4, which I factored as (x - 1)(x - 2)². So, the roots are x=1 and x=2 (with multiplicity 2). So, the function is undefined at x=1 and x=2.Therefore, the domain is all real numbers except x=1 and x=2, which in interval notation is (-∞, 1) ∪ (1, 2) ∪ (2, ∞).I think that's it. Let me just recap: to find the domain of a rational function, set the denominator equal to zero, solve for x, and exclude those values from the real numbers. In this case, the denominator factors into (x - 1)(x - 2)², so x=1 and x=2 are excluded. The numerator doesn't share these roots, so there are no holes, just vertical asymptotes at those points. Therefore, the domain is all real numbers except 1 and 2, expressed as the union of intervals I mentioned earlier.

💡First, I need to determine twice the number of cans of corn that Beth bought. Since she bought 10 cans of corn, twice that amount would be 2 multiplied by 10, which equals 20 cans.Next, I know that Beth bought 35 cans of peas. To find out how many more cans of peas she bought compared to twice the number of cans of corn, I subtract the doubled number of corn cans from the number of pea cans.So, 35 cans of peas minus 20 cans of corn equals 15 more cans of peas.Therefore, Beth bought 15 more cans of peas than twice the number of cans of corn.

💡Okay, so I have this geometry problem here. There's a circle and a square with the same center O, and they have equal areas. The circle has a radius of 1, and it intersects one side of the square at points P and Q. I need to find the length of PQ. The options are given as (A) √(4−π), (B) 1, (C) √2, (D) 2−√π, and (E) 4−√π.Alright, let me try to visualize this. There's a circle with radius 1, so its area is π*(1)^2 = π. The square has the same area as the circle, so the area of the square is also π. Since the area of a square is side length squared, the side length of the square must be √π. That makes sense because (√π)^2 = π.Now, the circle is centered at O, and so is the square. The circle intersects one side of the square at points P and Q. I need to find the distance between P and Q. Hmm, okay. So, PQ is a chord of the circle, right? Because it's a line segment connecting two points on the circumference of the circle.Since both the circle and the square are centered at O, the square is symmetric around O. So, the side of the square that the circle intersects must be at a certain distance from the center. Let me think about this.The square has a side length of √π, so half of that is (√π)/2. That would be the distance from the center O to the midpoint of any side of the square. So, if I consider the side where P and Q are located, the midpoint M of PQ is also the midpoint of that side of the square. Therefore, the distance from O to M is (√π)/2.Now, in the circle, the distance from the center O to the chord PQ is OM, which is (√π)/2. I remember that in a circle, the length of a chord can be found if we know the radius and the distance from the center to the chord. The formula for the length of a chord is 2*√(r^2 - d^2), where r is the radius and d is the distance from the center to the chord.So, in this case, the radius r is 1, and the distance d is (√π)/2. Plugging these into the formula, the length of PQ should be 2*√(1^2 - ((√π)/2)^2). Let me compute that step by step.First, compute ((√π)/2)^2. That would be (π)/4. Then, subtract that from 1: 1 - (π)/4. So, the expression under the square root becomes 1 - π/4. Then, take the square root of that, and multiply by 2.So, PQ = 2*√(1 - π/4). Let me simplify this expression. 1 - π/4 is the same as (4 - π)/4. So, √(1 - π/4) is √((4 - π)/4), which simplifies to √(4 - π)/2. Then, multiplying by 2, we get 2*(√(4 - π)/2) = √(4 - π).Wait, that's one of the answer choices. Option (A) is √(4−π). So, that must be the answer. Let me double-check my steps to make sure I didn't make a mistake.1. Area of the circle: π*(1)^2 = π.2. Area of the square: π, so side length is √π.3. Distance from center to the side of the square: (√π)/2.4. Using the chord length formula: 2*√(r^2 - d^2) = 2*√(1 - (π/4)) = 2*√((4 - π)/4) = √(4 - π).Yes, that seems correct. I think I got it right. So, the length of PQ is √(4−π), which is option (A).**Final Answer**boxed{A}

💡Okay, so I have this problem where I need to find the range of the real number ( m ) such that the intersection of sets ( A ) and ( B ) is equal to ( B ). That is, ( A cap B = B ). First, let me understand what this means. If ( A cap B = B ), that implies that every element of ( B ) is also in ( A ). In other words, ( B ) is a subset of ( A ). So, my goal is to find all values of ( m ) for which the set ( B ) is entirely contained within set ( A ).Alright, let me start by figuring out what set ( A ) is. The set ( A ) is defined as ( { x mid x^2 - 3x - 10 leq 0 } ). This is a quadratic inequality. To solve this, I need to find the values of ( x ) that satisfy the inequality ( x^2 - 3x - 10 leq 0 ).I remember that to solve a quadratic inequality, I should first find the roots of the corresponding quadratic equation ( x^2 - 3x - 10 = 0 ). Once I have the roots, I can determine the intervals where the quadratic expression is less than or equal to zero.So, let's solve ( x^2 - 3x - 10 = 0 ). I can use the quadratic formula here: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -3 ), and ( c = -10 ).Plugging in the values, I get:[x = frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 1 cdot (-10)}}{2 cdot 1} = frac{3 pm sqrt{9 + 40}}{2} = frac{3 pm sqrt{49}}{2} = frac{3 pm 7}{2}]So, the roots are:[x = frac{3 + 7}{2} = 5 quad text{and} quad x = frac{3 - 7}{2} = -2]Therefore, the solutions to the equation are ( x = 5 ) and ( x = -2 ).Now, to solve the inequality ( x^2 - 3x - 10 leq 0 ), I need to determine where the quadratic expression is less than or equal to zero. Since the quadratic opens upwards (because the coefficient of ( x^2 ) is positive), the expression will be less than or equal to zero between the two roots.So, the solution to the inequality is ( -2 leq x leq 5 ). Therefore, set ( A ) is the interval ( [-2, 5] ).Next, let's look at set ( B ). It is defined as ( { x mid m + 1 leq x leq 2m - 1 } ). So, ( B ) is an interval whose endpoints depend on the value of ( m ). The lower bound is ( m + 1 ) and the upper bound is ( 2m - 1 ).Since ( B ) is a subset of ( A ), the entire interval ( B ) must lie within ( A ). That means both endpoints of ( B ) must satisfy the conditions of being within ( A ). So, the lower bound ( m + 1 ) must be greater than or equal to ( -2 ), and the upper bound ( 2m - 1 ) must be less than or equal to ( 5 ).Additionally, for ( B ) to be a valid interval, the lower bound must be less than or equal to the upper bound. So, ( m + 1 leq 2m - 1 ). If this inequality is not satisfied, then ( B ) would be an empty set, which is trivially a subset of ( A ).Let me write down these conditions:1. ( m + 1 leq 2m - 1 ) (to ensure ( B ) is non-empty or empty)2. ( m + 1 geq -2 ) (lower bound of ( B ) must be within ( A ))3. ( 2m - 1 leq 5 ) (upper bound of ( B ) must be within ( A ))Let me solve each of these inequalities step by step.Starting with condition 1:[m + 1 leq 2m - 1]Subtract ( m ) from both sides:[1 leq m - 1]Add 1 to both sides:[2 leq m]So, ( m geq 2 ).Condition 2:[m + 1 geq -2]Subtract 1 from both sides:[m geq -3]Condition 3:[2m - 1 leq 5]Add 1 to both sides:[2m leq 6]Divide both sides by 2:[m leq 3]So, from conditions 2 and 3, we have ( -3 leq m leq 3 ). But from condition 1, we have ( m geq 2 ). Combining these, the range of ( m ) is ( 2 leq m leq 3 ).However, I should also consider the case where ( B ) is empty. If ( B ) is empty, then ( A cap B = B ) is trivially true because the intersection of any set with the empty set is the empty set. So, when is ( B ) empty? That happens when the lower bound is greater than the upper bound, which is when ( m + 1 > 2m - 1 ).Solving ( m + 1 > 2m - 1 ):[m + 1 > 2m - 1]Subtract ( m ) from both sides:[1 > m - 1]Add 1 to both sides:[2 > m]So, ( m < 2 ).Therefore, when ( m < 2 ), ( B ) is empty, and hence ( A cap B = B ) is automatically satisfied.Combining both cases:- When ( m < 2 ), ( B ) is empty, so the condition holds.- When ( 2 leq m leq 3 ), ( B ) is non-empty and a subset of ( A ), so the condition holds.Therefore, the range of ( m ) is all real numbers less than or equal to 3. In other words, ( m leq 3 ).Wait, hold on. Let me double-check that. If ( m ) is less than 2, ( B ) is empty, which is fine. If ( m ) is between 2 and 3, ( B ) is a valid interval within ( A ). But what if ( m ) is greater than 3? Then, the upper bound ( 2m - 1 ) would exceed 5, which is the upper limit of ( A ). So, ( B ) would not be entirely within ( A ), violating the condition ( B subseteq A ).Similarly, if ( m ) is less than -3, the lower bound ( m + 1 ) would be less than -2, which is the lower limit of ( A ). Thus, ( B ) would not be entirely within ( A ) either.But wait, in our earlier analysis, we found that for ( m geq 2 ), ( m ) must also be ( leq 3 ). So, combining the two cases, ( m leq 3 ) is the correct range because:- For ( m < 2 ), ( B ) is empty, which is acceptable.- For ( 2 leq m leq 3 ), ( B ) is a valid interval within ( A ).- For ( m > 3 ), ( B ) exceeds ( A ), which is not acceptable.Therefore, the range of ( m ) is all real numbers ( m ) such that ( m leq 3 ).I think that makes sense. Let me just recap:1. Solve for ( A ): ( A = [-2, 5] ).2. Define ( B ) as ( [m + 1, 2m - 1] ).3. For ( B subseteq A ), either ( B ) is empty or ( B ) is within ( A ).4. ( B ) is empty when ( m < 2 ).5. ( B ) is within ( A ) when ( 2 leq m leq 3 ).6. Therefore, combining both cases, ( m leq 3 ).Yes, that seems consistent. I don't think I missed anything here. So, the range of ( m ) is ( m leq 3 ).**Final Answer**The range of the real number ( m ) is boxed{m leq 3}.

💡First, I need to determine the total number of positive integers less than or equal to 36, which is simply 36.Next, I'll find all the divisors of 36. To do this, I'll start by performing the prime factorization of 36:36 = 2^2 cdot 3^2.Any divisor of 36 can be expressed in the form (2^a cdot 3^b), where (0 leq a leq 2) and (0 leq b leq 2). This means there are 3 possible values for (a) (0, 1, 2) and 3 possible values for (b) (0, 1, 2). Multiplying the number of choices for (a) and (b) gives the total number of divisors:3 times 3 = 9.Finally, the probability that a randomly chosen positive integer from 1 to 36 is a divisor of 36 is:frac{9}{36} = frac{1}{4}.

💡First, I need to determine the total number of possible three-digit numbers that can be formed using the digits 5, 6, and 7. Since each digit can be used only once, there are 3 choices for the first digit, 2 remaining choices for the second digit, and 1 choice for the third digit. This gives a total of 3 × 2 × 1 = 6 possible numbers.Next, I need to identify which of these numbers are common multiples of 3 and 5. A number is a multiple of 3 if the sum of its digits is divisible by 3, and it is a multiple of 5 if its last digit is either 0 or 5. Since we are only using the digits 5, 6, and 7, the last digit must be 5 to satisfy the multiple of 5 condition.Now, I will list all possible three-digit numbers and check which ones meet both criteria:- 567: Sum of digits = 5 + 6 + 7 = 18 (divisible by 3), but the last digit is 7 (not 5).- 576: Sum of digits = 5 + 7 + 6 = 18 (divisible by 3), but the last digit is 6 (not 5).- 657: Sum of digits = 6 + 5 + 7 = 18 (divisible by 3), but the last digit is 7 (not 5).- 675: Sum of digits = 6 + 7 + 5 = 18 (divisible by 3), and the last digit is 5.- 756: Sum of digits = 7 + 5 + 6 = 18 (divisible by 3), but the last digit is 6 (not 5).- 765: Sum of digits = 7 + 6 + 5 = 18 (divisible by 3), and the last digit is 5.From this analysis, only 675 and 765 are common multiples of 3 and 5. Therefore, there are 2 favorable outcomes out of 6 possible numbers.Finally, the probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes, which is 2/6. Simplifying this fraction gives a probability of 1/3.