answer:Let's denote the number of weeks in a month as W. Cathy was supposed to work 20 hours per week for 2 months, which is 2W weeks. So, without any changes, she would have worked 20 hours/week * 2W weeks = 40W hours. However, Cathy also took Chris's shifts for one week when Chris got sick. This means Cathy worked an extra 20 hours that week. We know that Cathy worked a total of 180 hours during the 2 months. So, we can set up the equation: 40W + 20 = 180 Now, let's solve for W: 40W = 180 - 20 40W = 160 W = 160 / 40 W = 4 So, there are boxed{4} weeks in a month.

answer:# Step-by-Step Solution Part (1): Finding the Range of Values for c Given that a, b, and c are the sides of triangle ABC and that a+b=3c-2 and a-b=2c-6, we can derive the conditions for c based on the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. 1. From a+b=3c-2, we know that the sum of two sides a and b is greater than the third side c, so we have: [3c-2 > c] Simplifying this inequality gives: [2c > 2] [c > 1] 2. From a-b=2c-6, considering the absolute value (since a-b could be negative, depending on which side is longer), we have: [|2c-6| < c] This inequality splits into two cases: - When 2c-6 geq 0, we have 2c-6 < c, which simplifies to c < 6. - When 2c-6 < 0, we have -(2c-6) < c, which simplifies to c > 3. Combining these conditions, we find that the range of values for c that satisfy both the triangle inequality and the given equations is 1 < c < 6. Therefore, the range of values for c is: [boxed{1 < c < 6}] Part (2): Finding the Value of c When the Perimeter is 18 Given that the perimeter of triangle ABC is 18, and knowing that a+b=3c-2, we can express the perimeter P as: [P = a + b + c = 3c - 2 + c = 4c - 2] Since the perimeter is 18, we have: [4c - 2 = 18] Solving for c gives: [4c = 20] [c = 5] Therefore, the value of c when the perimeter of triangle ABC is 18 is: [boxed{c = 5}]

answer:Given the points ( O(0,0,0) ), ( A(1,0,0) ), ( B(0,1,0) ), and ( C(0,0,1) ) in a 3-dimensional right-angle coordinate system, we are required to find the number of points that are equidistant from the planes ( OAB ), ( OBC ), ( OAC ), and ( ABC ). 1. **Identifying the Planes Equations**: - Plane ( OAB ): [ text{Passes through points } O, A, B. quad text{Equation} x = 0 ] - Plane ( OBC ): [ text{Passes through points } O, B, C. quad text{Equation} y = 0 ] - Plane ( OAC ): [ text{Passes through points } O, A, C. quad text{Equation} z = 0 ] - Plane ( ABC ): [ text{Plane passing through points } A, B, C. quad text{Equation} x + y + z = 1 ] 2. **Distance Calculation**: For a point ( P(x, y, z) ) to be equidistant from all four planes: - Distance from ( P ) to the plane ( OAB ) (i.e., plane ( x = 0 )) is (|x|). - Distance from ( P ) to the plane ( OBC ) (i.e., plane ( y = 0 )) is (|y|). - Distance from ( P ) to the plane ( OAC ) (i.e., plane ( z = 0 )) is (|z|). - Distance from ( P ) to the plane ( ABC ) (i.e., plane ( x + y + z = 1 )) is [ left|frac{x + y + z - 1}{sqrt{1^2 + 1^2 + 1^2}}right| = frac{|x + y + z - 1|}{sqrt{3}} ] 3. **Setting Up Equidistant Conditions**: Let's set the condition for equality: [ |x| = |y| = |z| = frac{|x + y + z - 1|}{sqrt{3}} ] 4. **Analyzing Possible Solutions**: Since (|x| = |y| = |z|), assume ( |x| = |y| = |z| = d ). So, [ |d + d + d - 1| = sqrt{3} d ] Simplifying, [ |3d - 1| = sqrt{3} d ] 5. **Solving Absolute Value Equation**: - Case 1: ( 3d - 1 = sqrt{3} d implies 3d - sqrt{3} d = 1 implies d (3 - sqrt{3}) = 1 implies d = frac{1}{3 - sqrt{3}} ) Rationalize the denominator: [ d = frac{1}{3 - sqrt{3}} cdot frac{3 + sqrt{3}}{3 + sqrt{3}} = frac{3 + sqrt{3}}{9 - 3} = frac{3 + sqrt{3}}{6} = frac{3(1 + 1/sqrt{3})}{6} = frac{1 + 1/sqrt{3}}{2} ] - Case 2: ( 3d - 1 = -sqrt{3} d implies 3d + sqrt{3} d = 1 implies d (3 + sqrt{3}) = 1 implies d = frac{1}{3 + sqrt{3}} ) Rationalize the denominator: [ d = frac{1}{3 + sqrt{3}} cdot frac{3 - sqrt{3}}{3 - sqrt{3}} = frac{3 - sqrt{3}}{9 - 3} = frac{3 - sqrt{3}}{6} = frac{1 - 1/sqrt{3}}{2} ] Combining the observations for equidistant points yields five distinct points: the four midpoints of ({0,0,d}), ({0,d,0}), ({d,0,0}) using the positive and negative solutions for (d), and the incenter of tetrahedron. **Conclusion**: [ boxed{5}

answer:Let's denote the cricketer's average score before the 19th inning as "A". This means that after 18 innings, the cricketer has scored a total of 18A runs. In the 19th inning, the cricketer scores 99 runs. This increases his average by 4, so his new average after 19 innings is A + 4. The total runs scored after 19 innings is the sum of the runs scored in the first 18 innings and the runs scored in the 19th inning, which is 18A + 99. We can now set up an equation using the new average (A + 4) and the total runs scored after 19 innings: 19(A + 4) = 18A + 99 Expanding the left side of the equation, we get: 19A + 76 = 18A + 99 Now, we'll solve for A: 19A - 18A = 99 - 76 A = 23 So, the cricketer's average score before the 19th inning was 23 runs. After scoring 99 runs in the 19th inning, his new average score after 19 innings is: A + 4 = 23 + 4 = 27 Therefore, the cricketer's average score after 19 innings is boxed{27} runs.