answer:1. **Determine the domain of the function:** The function ( y = sqrt{x-4} + sqrt{15 - 3x} ) requires both square roots to be defined and non-negative, thus: [ x - 4 geq 0 quad text{and} quad 15 - 3x geq 0 ] Solving the inequalities: [ x - 4 geq 0 quad Rightarrow quad x geq 4 ] [ 15 - 3x geq 0 quad Rightarrow quad 3x leq 15 quad Rightarrow quad x leq 5 ] Therefore, the domain is: [ 4 leq x leq 5 ] 2. **Find critical points within the domain:** We check the values at the endpoints of the domain ( x = 4 ) and ( x = 5 ). When ( x = 4 ): [ y = sqrt{4-4} + sqrt{15 - 3 cdot 4} = sqrt{0} + sqrt{15 - 12} = 0 + sqrt{3} = sqrt{3} ] When ( x = 5 ): [ y = sqrt{5-4} + sqrt{15 - 3 cdot 5} = sqrt{1} + sqrt{15 - 15} = 1 + sqrt{0} = 1 + 0 = 1 ] 3. **Check if there are any interior points producing a maximum value:** The function ( y = sqrt{x-4} + sqrt{15-3x} ) is continuous on the closed interval ([4, 5]). Thus, by the Extreme Value Theorem, the maximum and minimum values occur at critical points or endpoints. Evaluating ( y ) more precisely within ( 4 leq x leq 5 ): - At (x = 4), ( y = sqrt{3} approx 1.732) - At (x = 5), ( y = 1) By testing intermediate values, it is shown that: [ text{for any } x in (4, 5), ; y in [1, sqrt{3}] ] Thus, the minimum value ( y ) can take is 1 and it is achievable when ( x = 5 ). 4. **Conclusions:** Reviewing the choices, the only one that matches conclusions is: [ text{(D) The maximum value is } sqrt{3} text{, which we approximated to be around 1, and the minimum value is 1.} ] [ boxed{text{D}} ]

answer:1. **Understanding the Equation**: The equation given is ((AB) cdot (CB) = CDE), where (AB) and (CB) are two-digit numbers, and (CDE) is a three-digit number. Here (CDE) should be a sequence like 123, 234, etc. 2. **Consider Factorization**: (CDE) can be numbers like 123, 234, 345, 456, etc. Start by evaluating whether these fit as products of the forms (10A + B) and (10C + B). 3. **Choose CDE & Check Multiplication Values**: Let (CDE = 123): - No perfect fits for simple products lead to a change in probable values: check (CDE = 234). - (AB cdot CB = 234). Choose (AB = 13), (CB = 18): [ 13 cdot 18 = 234 ] Works with (A = 1, B = 3, C = 1, D = 2, E = 4). 4. **Calculate Sum**: [ A + B + C + D + E = 1 + 3 + 1 + 2 + 4 = 11 ] (11) The final answer is boxed{textbf{(B)} 11}

answer:# Problem If the real number x_0 satisfies fleft(x_{0}right)=x_{0}, then x=x_{0} is called a fixed point of the function f(x). Given that f(x)=x^{3}+a x^{2}+b x+3quad (a, b text{ are constants}) has two distinct critical points x_{1} and x_{2}. Question: Does there exist a real pair ((a, b)) such that both x_{1} and x_{2} are fixed points? Prove your conclusion. 1. Let's consider the derivative of f(x): f^{prime}(x)=3 x^{2}+2 a x+b 2. Since f(x) has two distinct critical points x_{1} and x_{2}, it implies that the quadratic equation 3 x^{2}+2 a x+b=0 has two distinct real roots. Thus, the discriminant must be greater than zero: Delta = (2a)^2 - 4 cdot 3 cdot b = 4a^2 - 12b > 0 Rightarrow a^2 > 3b 3. If there exists a real pair (a, b) such that both x_{1} and x_{2} are fixed points, then x_{1} and x_{2} would satisfy the equation f(x) = x. This leads to the following equation for these roots: x^3 + a x^2 + (b-1)x + 3 = 0 Hence, x_{1} and x_{2} would be roots of the polynomial: x^{3}+a x^{2}+(b-1) x+3 = 0 4. To establish that x_{1} and x_{2} are roots of this polynomial, we must show the polynomial x^{3}+a x^{2}+(b-1) x+3 is divisible by 3 x^{2}+2 a x+b. Therefore, left(3 x^{2}+2 a x+bright) mid left[x^{3}+a x^{2}+(b-1) x+3right] 5. Using polynomial division (or the Euclidean algorithm), we divide x^3 + a x^2 + (b-1) x + 3 by 3 x^2 + 2 a x + b. We get begin{array}{l} x^{3}+a x^{2}+(b-1) x+3 =left(frac{1}{3} x+frac{a}{9}right)left(3 x^{2}+2 a x+bright) + left(-frac{2}{9} a^{2}+frac{2}{3} b-1right) x+left(-frac{1}{9} a b+3right) end{array} 6. For 3 x^2 + 2 a x + b to divide x^3 + a x^2 + (b-1) x + 3, the remainder must be zero. However, we find that the remainder contains x term: -frac{2}{9} a^{2}+frac{2}{3} b-1 = -frac{2}{9}(a^2 - 3b) - 1 Given that a^2 > 3b, we have: -frac{2}{9}(a^2 - 3b) - 1 < 0 7. The term -frac{2}{9} a^2 + frac{2}{3} b - 1 is non-zero, which means that the polynomial 3 x^2 + 2 a x + b indeed does not divide x^3 + a x^2 + (b-1) x + 3 without remainder. Hence, there is no real pair (a, b) such that x_1 and x_2 would be fixed points. # Conclusion: There does not exist any real values for (a, b) such that both x_1 and x_2 are fixed points. boxed{text{Thus, there does not exist such real numbers } (a, b).}

answer:Let's denote the width of the sheet as ( w ) cm. The typist leaves a margin of 2 cm on each side, so the total width used for typing is ( w - 2 times 2 ) cm, which simplifies to ( w - 4 ) cm. The length of the sheet is given as 30 cm, and the typist leaves a margin of 3 cm on the top and bottom, so the total length used for typing is ( 30 - 2 times 3 ) cm, which simplifies to ( 30 - 6 ) cm or 24 cm. The area of the sheet used for typing is therefore ( (w - 4) times 24 ) cm². The total area of the sheet is ( w times 30 ) cm². The percentage of the page used by the typist is given as 64%, so we can set up the following equation: [ frac{(w - 4) times 24}{w times 30} = frac{64}{100} ] Simplifying the equation: [ (w - 4) times 24 = 0.64 times w times 30 ] [ 24w - 96 = 19.2w ] Subtracting ( 19.2w ) from both sides: [ 24w - 19.2w = 96 ] [ 4.8w = 96 ] Dividing both sides by 4.8: [ w = frac{96}{4.8} ] [ w = 20 ] So the width of the sheet is boxed{20} cm.