answer:Let's denote the number of fish Ken caught as K. According to the information given, Ken released 3 fish back into the lake, so the number of fish he brought home is K - 3. Kendra caught 30 fish and did not release any, so she brought home all 30 fish. Together, Ken and Kendra brought home 87 fish. Therefore, we can write the following equation: (K - 3) + 30 = 87 Now, let's solve for K: K - 3 = 87 - 30 K - 3 = 57 K = 57 + 3 K = 60 So, Ken caught 60 fish in total, but since he released 3, he brought home 60 - 3 = 57 fish. Now, we can find the ratio of the number of fish Ken caught to the number of fish Kendra caught: Ken's fish : Kendra's fish = 57 : 30 To simplify the ratio, we can divide both numbers by their greatest common divisor, which is 3: 57 ÷ 3 : 30 ÷ 3 = 19 : 10 Therefore, the ratio of the number of fish Ken caught to the number of fish Kendra caught is boxed{19:10} .

answer:1. **Define Key Points and Cyclic Quadrilaterals:** Let ( M ) be the midpoint of (overline{BH}) and ( B_1 ) be the foot of the perpendicular from ( B ) to (overline{AC}). Since ( AA_1 ) and ( CC_1 ) are altitudes, ( H ) is the orthocenter of (triangle ABC). The quadrilaterals ( AB_1C_1H ) and ( CB_1A_1H ) are cyclic because the opposite angles are right angles. 2. **Establish Collinearity:** We need to show that ( M, C_1, X ) and ( M, A_1, Y ) are collinear. This will imply that ( X ) and ( Y ) are equidistant from ( M ) because ( M ) is the midpoint of (overline{BH}) and lies on the perpendicular bisector of (overline{C_1H}) and (overline{A_1H}). 3. **Define ( X' ) and ( Y' ):** Define ( X' = overline{MC_1} cap (AB_1C_1H) neq C_1 ). Since ( M ) lies on the perpendicular bisector of (overline{C_1H}), the quadrilateral ( B_1HC_1X' ) is an isosceles trapezoid. This implies that ( angle C_1HX' = angle B_1XH ). 4. **Angle Chasing:** We have: [ angle B_1XH = angle B_1AH = angle HC_1A_1 ] This shows that ( overline{X'H} parallel overline{A_1C_1} ). 5. **Define ( Y' ) Similarly:** Define ( Y' ) similarly such that ( Y' = overline{MA_1} cap (CB_1A_1H) neq A_1 ). By the same reasoning, ( overline{Y'H} parallel overline{A_1C_1} ). 6. **Collinearity of ( X', H, Y' ):** Since both ( overline{X'H} ) and ( overline{Y'H} ) are parallel to ( overline{A_1C_1} ), it follows that ( X', H, Y' ) are collinear and ( overline{X'HY'} parallel overline{A_1C_1} ). 7. **Conclusion:** Therefore, ( X' = X ) and ( Y' = Y ), proving that ( X ) and ( Y ) are equidistant from ( M ). (blacksquare)

answer:Let's call the number "x". According to the problem, 70% of the number x is calculated and then 40 is subtracted from it to get a certain result. First, we find 70% of x: 70% of x = (70/100) * x Since we know the number x is 100, we can substitute x with 100: 70% of 100 = (70/100) * 100 = 70 Now, we subtract 40 from this value to get the result: Result = 70 - 40 = 30 So, the result is boxed{30} .

answer:The factorial 8! = 40320 can be factored as 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2. To find w, x, y, and z, it's effective to group these factors into four products where the values are close together and the order w < x < y < z is maintained. Performing similar analysis as with the three factors, we find: - We can take w = 7 cdot 2 = 14 - For x, using 6 cdot 4 = 24 seems suitable - Then taking y as 5 cdot 3 = 15 - And finally, z = 8 Thus, the assignment (w, x, y, z) = (14, 24, 15, 8). However, the ordering w < x < y < z is not maintained. Under correct order, w = 8, x = 14, y = 15, z = 24. The difference z - w = 24 - 8. Therefore, the smallest possible value of z - w is boxed{16}.