Quadratics Applications Essay

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Quadratics Applications

If 221 square Inches of picture shows, find the width of the frame, assuming the distance is equal all the way around. Let of frame A-law EX.-XX+31=o x=1 5. 5 x=O. 5 15. 5 is not possible. That is larger than 14, one of the sides of the frame. The frame has a width of 0. 5 inches. 3. A bulging architect Is modifying the visual space of a bulging with mirrors. One base of the mirror. If the mirror has an area of 108 square feet, what are the base and height of the mirror? Let x=base and xx-6=height A=be 216=EX.-XX EX.-xx-216=o x=12 -9 The base of the mirror must be 12 feet. Since -9 does not make sense, the height is 8 feet. . The length of the side of a large square is 1 CM less than twice the length of the side of a smaller square. The area of the large square is 33 com more than the area of the small square. Find the length of the sides of the two squares. Let x=side of small square xx-1=side of large square AL=AS+33 ex.-xx+ I ”x-2*33 EX.-xx-32=o The small square has a side of 4 CM since -2 does not make sense. The large square has side lengths of 7 CM. 5. The perimeter off rectangle is 36 m. The area is 65 mm. Find the dimensions of the rectangle. P=2(l+w) 18=1+w Let width be x. 65=XX-XX XX+65=o Then dimensions are x and 18-x.

Part 2: Vertical Motion Formula x=13 For an object thrown into the air with an initial upward velocity of and an initial height of , then its distance from the surface is found by the following formulas. T represents seconds in the air. D represents distance in meters from the ground. D represents distance in feet from the ground. 6. Felix is playing in a soccer game. He kicks the soccer ball into the air with an initial upward velocity of 25 meters per second (m/sec) with an initial height at ground level. A. Calculate the height of the soccer ball after 2 seconds and 3 seconds. (2) = + 25(2) + o (2) = 30 m d(3) = + 25(3) + O d(3) = 30 m b. When will the soccer ball be 20 meters above the ground? 20=-5th+25th+0 0=-5th+25th-20 It will be 20 meters above the ground at 1 second and 4 seconds. C. Sketch a graph of the function using an appropriate domain and range. D. When will the soccer ball hit the ground? By looking at the root of the graph, the ball will hit the ground 5 seconds after it’s kicked. E. What will be the greatest height reached by the soccer ball and when will this occur? How can this be determined from the graph? The greatest height is 31. 25 Ft. , which is 2. 5 sec. After it’s kicked.

This is represented by the maximum point, or vertex, of the parabola. Per second. A. How high is the skeet after 0. 5 seconds, 1 second, 3 seconds? D = -16th +96th; 144 Ft b. If the shooter hits the skeet when it has been in the air 5 seconds, how high is the skeet? D(5) = 80 Ft c. If the skeet is 128 feet in the air when hit by the shooter, what are the possible times it was in the air? 4 sec, 2 sec 128=-16th+96th+0 0=-16th+96th-128 It will be 128 feet above the ground at 2 seconds and 4 seconds. D. If the shooter never hits the skeet, when will it hit the ground? -16th + 96th It will hit the ground after 6 seconds. Sketch a graph of the function with an appropriate window. 8. A rock is thrown upward from a 50-meter tall cliff on the water edge at an initial velocity of 15 m/s. A. How high will the rock be from ground level after 2 seconds? D – 15th+50; b. When will it be at the same height as when it was first thrown? -5th+ 15th+ 50 t = O t = 3 It will be at the same height as thrown at 3 seconds. C. How high will the rock be from ground level after 4 seconds? Where is this in relationship to the top of the cliff? D(4) = 30 m. The rock will be 20 meters below the top of the cliff. D. When will the rock hit the water?

O = -5th + 15th + 50. The rock hits the water at 5 seconds. 9. One contest in the winter Olympics is ski Jumping. A contestant makes a Jump with an initial upward velocity of 13 m/s and an initial height at the end of the ski Jump ramp of 60 meters. A. How high will the contestant be after 1 second? D = -5th + 13th + 60; m b. After two seconds will the contestant be going up or down? Explain. Down because the parabola is turning downward at two seconds. C. What is the maximum height the ski Jumper will reach and at what elapsed time will this occur? The Jumper will reach 68. 45 meters at 1. 3 seconds. D.

When will the contestant again be at the same height as when he left the end of the ski Jump ramp? 60=-5th+ 13th+60; t = O t = 2. 6 The contestant will be at the same height at 2. 6 seconds. E. When will the contestant land the Jump? O = -5th + 13th + 60; The contestant lands the Jump at 5 seconds. F. Sketch a graph of the function using an appropriate domain and range. 10. On naval battleships large guns must fire rockets many kilometers. The rockets are projected with both upward and horizontal direction. From one large gun, rockets are fired with an initial velocity of 300 m/s and an approximate initial height of O teeters. . Twenty seconds after it is fired, what will be the height of the rocket? D -5th + soot; d(20) = 4000 m b. When will it reach a height of 1 kilometer? 1 km=1000 m; 1000 = -5th + 30th. The rocket reaches the height of 1 kilometer at about 3. 54 seconds. C. Will the rocket ever attain a height of 5000 meters? Explain. No, because the maximum height the rocket reaches is 4500 meters. D. How long does it take to reach a target at ground level? O = -5th + 30th. It takes 60 seconds to reach a target at ground level. E. Sketch a graph of the function using an appropriate domain and range. 1 1 .

A diver practices Jumping from a trampoline that is 9 feet above a cushioned surface. The initial velocity of the Jump is 12 Ft. ‘s. A. When will the diver again be at the same level as the height from which he notionally jumped? D=-16th+ 12th+9; 0. 75 sec b. What will be the maximum height reached by the diver and at what elapsed time will this occur? The diver will reach maximum height of 1 1. 25 feet at 0. 37 seconds. C. When will the diver reach the cushioned surface? O = -16th+ 12th + 9. The diver will reach the cushioned surface at 1. 21 seconds. If the diver were on the moon, gravity would be much weaker.

The equation would be: Suppose the initial velocity of the Jump is still 12 Ft. ‘s. D. When will the diver again be at the same level as the height from which he originally Jumped on the moon? D = -2. 4th + 12th+9; 9= -2. 4th + 12th + 9. The diver will be at the same height at 5 seconds. E. What will be the maximum height reached by the diver and at what elapsed time will this occur on the moon? The diver would reach the maximum height of 24 Ft at 2. 5 seconds. F. When will the diver reach the cushioned surface on the moon? O The diver will reach the cushioned surface at 5. 66 seconds. = -2. 4th+12th+9.