answer:To find the greatest two-digit prime saturated integer, we need to consider the definition of a prime saturated number. A number ( z ) is prime saturated if the product of all the different positive prime factors of ( z ) is less than the square root of ( z ). Let's denote the product of all the different positive prime factors of ( z ) as ( P ). Then, according to the definition, we must have: [ P < sqrt{z} ] Since we are looking for the greatest two-digit prime saturated integer, let's start from the largest two-digit number, which is 99, and work our way down until we find a number that satisfies the condition. The prime factors of 99 are 3 and 11, and their product is ( P = 3 times 11 = 33 ). The square root of 99 is approximately 9.95. Since ( P > sqrt{z} ), 99 is not prime saturated. Let's try the next number, 98. The prime factors of 98 are 2 and 7, and their product is ( P = 2 times 7 = 14 ). The square root of 98 is approximately 9.90. Since ( P < sqrt{z} ), 98 is prime saturated. We have found a two-digit prime saturated integer, 98. However, we need to check if there is a larger two-digit prime saturated integer. The next number, 97, is a prime number, so its only prime factor is 97 itself, and ( P = 97 ). The square root of 97 is approximately 9.85. Since ( P > sqrt{z} ), 97 is not prime saturated. We can stop here because any number less than 97 will have a smaller product of prime factors or a larger square root, making it less likely to be prime saturated than 98. Therefore, the greatest two-digit prime saturated integer is boxed{98} .

answer:First, we find the prime factorization of 60: [ 60 = 2^2 cdot 3 cdot 5. ] The number 60 can have: - zero or two powers of 2, - zero or one power of 3, - zero or one power of 5. To find the number of factors, we add one to each of the exponents in the prime factorization and multiply these results together: [ (2+1)cdot(1+1)cdot(1+1) = 3 cdot 2 cdot 2 = 12. ] Therefore, there are boxed{12} factors of 60. These factors, verified by direct computation or listing, are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

answer:To show that a triangle is right-angled if and only if (2R = s - r), we will use some known properties of triangles and their circumradius (R), inradius (r), and semi-perimeter (s). 1. **Express the semi-perimeter (s), circumradius (R), and inradius (r):** - The semi-perimeter (s) of a triangle with sides (a), (b), and (c) is given by: [ s = frac{a + b + c}{2} ] - The circumradius (R) of a triangle is given by: [ R = frac{abc}{4K} ] where (K) is the area of the triangle. - The inradius (r) of a triangle is given by: [ r = frac{K}{s} ] 2. **Use the relationship (2R = s - r):** - We need to show that (2R = s - r) if and only if the triangle is right-angled. 3. **Consider the area (K) of the triangle:** - For a right-angled triangle, the area (K) can be expressed as: [ K = frac{1}{2} times text{leg}_1 times text{leg}_2 ] where (text{leg}_1) and (text{leg}_2) are the two legs of the right-angled triangle. 4. **Express (K) in terms of (a), (b), and (c):** - For a right-angled triangle with hypotenuse (c) and legs (a) and (b), the area (K) is: [ K = frac{1}{2}ab ] 5. **Substitute (K) into the formulas for (R) and (r):** - The circumradius (R) for a right-angled triangle is: [ R = frac{abc}{4K} = frac{abc}{4 times frac{1}{2}ab} = frac{c}{2} ] - The inradius (r) for a right-angled triangle is: [ r = frac{K}{s} = frac{frac{1}{2}ab}{frac{a + b + c}{2}} = frac{ab}{a + b + c} ] 6. **Verify the relationship (2R = s - r):** - For a right-angled triangle: [ 2R = 2 times frac{c}{2} = c ] - The semi-perimeter (s) is: [ s = frac{a + b + c}{2} ] - The inradius (r) is: [ r = frac{ab}{a + b + c} ] - Therefore: [ s - r = frac{a + b + c}{2} - frac{ab}{a + b + c} ] - For a right-angled triangle, we have: [ c = sqrt{a^2 + b^2} ] - Substituting (c) into the equation: [ s - r = frac{a + b + sqrt{a^2 + b^2}}{2} - frac{ab}{a + b + sqrt{a^2 + b^2}} ] - Simplifying, we get: [ s - r = c ] - Hence, (2R = s - r) holds true for a right-angled triangle. 7. **Conversely, if (2R = s - r), then the triangle must be right-angled:** - If (2R = s - r), then: [ c = frac{a + b + c}{2} - frac{ab}{a + b + c} ] - This implies that the triangle satisfies the Pythagorean theorem, and thus it must be right-angled. Therefore, we have shown that a triangle is right-angled if and only if (2R = s - r). (blacksquare)

answer:1. **Finding the Minimum Point of the Original Graph:** Since |x| is non-negative, it is minimized when x = 0. Substituting x=0 into the original equation: [ y = 2|0| - 3 = 0 - 3 = -3 ] Therefore, the minimum point of the graph of y = 2|x| - 3 is at (0, -3). 2. **Applying the Translation:** The graph is translated four units to the right, which adds four to the x-coordinate of the minimum point, and one unit up, which adds one to the y-coordinate. Therefore, the new coordinates of the minimum point are: [ (0 + 4, -3 + 1) = (4, -2) ] So, the coordinates of the new minimum point after the translation are boxed{(4, -2)}.