ARRANGE:B. a= kb/hD. 1/2F. Z=kx/yU. 6N. a=kbhA. 44 cm²H. z=kxyanswer: _ /1 ,_ /2, _ /3, _ /4,_ /5,_ /6➤ Activity 19: Think Deeper!Solve the following:1. The amount of gasoline used by a car varies jointly as the distance travelled and the square root of the speed. Suppose a car used 25 liters on a 100 kilometer trip at 100 km/hr. About how many liters will it use on a 192 kilometer trip at 64 km/hr?2. The area of triangle varies jointly as the base and the height. A triangle with a base of 8 cm and a height of 9 cm has an area of 36 square centimeters. Find the area when the base is 10 cm and the height is 7 cm.3. The volume of a right circular cylinder varies jointly as the height and the square of the radius. The volume of a right circular cylinder, with radius 4 centimeters and height 7 centimeters is 112 cm³. Find the volume of another cylinder with radius 8 centimeters and height 14 centimeters.4. The mass of a rectangular sheet of wood varies jointly as the length and the width. When the length is 20 cm and the width is 10 cm, the mass is 200 grams. Find the mass when the length is 15 cm and the width is 10 cm.5. The weight of a rectangular block of metal varies jointly as its length, width and thickness. If the weight of a 13 dm by 8 dm by 6 dm block of aluminum is 18.2 kg, find the weight of a 16 dm by 10 dm by 4 dm block of aluminum.
Answer (1)
Answer:Problem 1: Gasoline UsageLet G = k × d × √s, where G = gasoline used, d = distance, s = speed, and k = constant.Given: G = 25 liters, d = 100 km, s = 100 km/hrFind k: 25 = k × 100 × √100k ≈ 0.025Now, find G for d = 192 km, s = 64 km/hr:G ≈ 0.025 × 192 × √64[tex] \large \text{G ≈ 38.4 liters}[/tex]Problem 2: Triangle AreaLet A = k × b × h, where A = area, b = base, h = height, and k = constant.Given: A = 36 cm², b = 8 cm, h = 9 cmFind k: 36 = k × 8 × 9k = 0.5Now, find A for b = 10 cm, h = 7 cm:A = 0.5 × 10 × 7[tex] \large \text{A = 35 cm²}[/tex]Problem 3: Cylinder VolumeLet V = k × h × r², where V = volume, h = height, r = radius, and k = constant.Given: V = 112 cm³, h = 7 cm, r = 4 cmFind k: 112 = k × 7 × 16k ≈ 1Now, find V for h = 14 cm, r = 8 cm:V ≈ 1 × 14 × 64[tex] \large \text{V ≈ 896 cm³}[/tex]Problem 4: Wood MassLet M = k × l × w, where M = mass, l = length, w = width, and k = constant.Given: M = 200 grams, l = 20 cm, w = 10 cmFind k: 200 = k × 20 × 10k = 1Now, find M for l = 15 cm, w = 10 cm:M = 1 × 15 × 10[tex] \large \text{M = 150 grams}[/tex]Problem 5: Metal WeightLet W = k × l × w × t, where W = weight, l = length, w = width, t = thickness, and k = constant.Given: W = 18.2 kg, l = 13 dm, w = 8 dm, t = 6 dmFind k: 18.2 = k × 13 × 8 × 6k ≈ 0.0385Now, find W for l = 16 dm, w = 10 dm, t = 4 dm:W ≈ 0.0385 × 16 × 10 × 4[tex] \large \text{W ≈ 24.64 kg}[/tex]
