answer:Since A={0,2,3} and B={x+1,x^{2}+4}, and A cap B = {3}, Either x+1=3 or x^{2}+4=3, Solving these equations, we get x=2 or no solution. Thus, the value of the real number x is 2. To find the value of x, we determine the intersection of sets A and B. This problem tests our understanding of the intersection operation and its definition. Proficiency in the definition of intersections is crucial to solving this problem. The final answer is: boxed{2}.

answer:1. **Domain of Definition (ODZ) Requirements**: To find when the function ( y = sqrt[4]{10+x} - sqrt{2-x} ) takes positive values, we first consider the domains of the component functions: - The fourth root, ( sqrt[4]{10+x} ), requires ( 10 + x geq 0 ) or ( x geq -10 ). - The square root, ( sqrt{2-x} ), requires ( 2 - x geq 0 ) or ( x leq 2 ). 2. **Function Positivity Requirement**: We need ( sqrt[4]{10+x} - sqrt{2-x} > 0 ). 3. **Combining Requirements**: We combine the domain requirements with the positivity condition: [ left{ begin{array}{l} sqrt[4]{10 + x} - sqrt{2 - x} > 0, x geq -10, x leq 2 end{array} right. ] 4. **Simplifying the Positivity Condition**: Suppose ( z = sqrt[4]{10 + x} ). Then ( z^4 = 10 + x ) and we get: [ z > sqrt{2-x} ] Squaring both sides (since both are non-negative): [ z^2 > 2 - x ] But, ( z^2 = sqrt{10 + x} ). Then: [ sqrt{10 + x} > 2 - x ] Squaring both sides again, we obtain: [ 10 + x > (2 - x)^2 ] Expanding and solving: [ 10 + x > 4 - 4x + x^2 ] Rearranging the inequality: [ x^2 - 5x - 6 < 0 ] 5. **Solving the Quadratic Inequality**: Let's find the roots of the quadratic equation ( x^2 - 5x - 6 = 0 ): [ x = frac{5 pm sqrt{25 + 24}}{2} = frac{5 pm 7}{2} ] The roots are: [ x = 6 quad text{and} quad x = -1 ] The quadratic inequality ( x^2 - 5x - 6 < 0 ) is satisfied between these roots: [ -1 < x < 6 ] 6. **Combining All Conditions**: We must also satisfy: [ x geq -10 quad text{and} quad x leq 2 ] Therefore, combining: [ -1 < x < 6 quad text{with} quad -10 leq x leq 2 ] Results in: [ -1 < x leq 2 ] # Conclusion: The function ( y = sqrt[4]{10 + x} - sqrt{2 - x} ) is positive for: [ boxed{-1 < x leq 2} ]

answer:Given the operation 5 bowtie x = 5 + sqrt{x + sqrt{x + sqrt{x + ...}}}, and it's given that this equals 11. Therefore, 5 + sqrt{x + sqrt{x + sqrt{x + ...}}} = 11. Isolating the nested radical, we have: sqrt{x + sqrt{x + sqrt{x + ...}}} = 11 - 5 = 6. Since the series of nested radicals continues indefinitely, we can replace the entire nested radical with 6, giving: sqrt{x + 6} = 6. Squaring both sides results in: x + 6 = 36. Solving for x, we find: x = 36 - 6 = boxed{30}.

answer:Let (BC=x), then (AC= sqrt{3}x), According to the area formula, we have (S_{triangle ABC}= frac{1}{2}AB cdot BC sin B) (= frac{1}{2} times 2x sqrt{1-cos^2 B}), According to the cosine theorem, we get (cos B= frac{4+x^2-3x^2}{4x}= frac{2-x^2}{2x}), Substituting into the above formula, we get (S_{triangle ABC}= frac{1}{2} sqrt{-(x^2-4)^2+12}), By the relationship of the three sides of a triangle, we have (begin{cases} sqrt{3}x+x > 2 x+2 > sqrt{3}x end{cases}), Solving this, we get (sqrt{3}-1 < x < sqrt{3}+1). Therefore, when (x=2), (S_{triangle ABC}) reaches its maximum value (sqrt{3}). Let (BC=x), use the area formula with (x) and (sin B) to express the area of the triangle, then use the cosine theorem with (x) to express (sin B), and substitute it into the expression for the area of the triangle, thereby obtaining an expression for the area of the triangle in terms of (x), and then find the maximum value of the triangle's area based on the range of (x). This problem mainly examines the application of the cosine theorem and the area formula in solving triangles. When dealing with maximum or minimum problems, consider using the monotonicity of functions and the domain of definition, etc. Therefore, the maximum area of triangle (ABC) is boxed{sqrt{3}}.