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The document discusses 2nd degree equations in the context of physics and geometry word problems. It provides an example of calculating the height, time, and maximum height of a stone thrown straight up using the equation h=-16t^2 + vt. It also discusses the area formula for a rectangle, A=LW, and gives an example problem to find the length and width of a rectangle given its area is 21 in^2 and the length is 4 inches more than the width.

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2nd-Degree-Equation Word Problems
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree.
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt where h = height in feet t = time in second v = upward speed in feet per second
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second?
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64,
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1)
2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1) h = -16 + 64 = 48 ft
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground?
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0.
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4)
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds.
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained?
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point.
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2)
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128 = 64 (ft)
2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128 = 64 (ft) Therefore, the maximum height is 64 feet.
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations.
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle,
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW.
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. If L and W are in a given unit, then A is in unit2. L w A = LW
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2 1 in 1 in 1 in2 L w A = LW
Formulas of area in mathematics also lead to 2nd degree equations. If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2 1 in 1 in 1 in2 2 in 3 in 6 in2 2nd-Degree-Equation Word Problems Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram L H=height B=base
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. L H=height B=base
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, L H=height B=base
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, H=height B=base L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we get a rectangle. H=height B=base L w A = LW
2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we get a rectangle. Hence the area A of the parallelogram is A = BH. H=height B=base L w A = LW
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. x 2x + 3
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 Area of a Triangle x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown .
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown.
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram.
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram. If A is the area of the triangle,
2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram. If A is the area of the triangle, then 2A = HB or A = B H . 2
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) x 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x x 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x x 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3
2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3 Therefore the height is 4 in. and the base is 5 in.
2nd-Degree-Equation Word Problems Exercise A. Use the formula h = –16t2 + vt for the following problems. 1. A stone is thrown upward at a speed of v = 64 ft/sec, how long does it take for it’s height to reach 48 ft? 2. A stone is thrown upward at a speed of v = 64 ft/sec, how long does it take for it’s height to reach 28 ft? 3. A stone is thrown upward at a speed of v = 96 ft/sec, a. how long does it take for its height to reach 80 ft? Draw a picture. b. how long does it take for its height to reach the highest point? c. What is the maximum height it reached? 4. A stone is thrown upward at a speed of v = 128 ft/sec, a. how long does it take for its height to reach 256 ft? Draw a picture. How long does it take for its height to reach the highest point and what is the maximum height it reached?
2nd-Degree-Equation Word Problems B. Given the following area measurements, find x. 5. 8 ft2 x + 2 x 6. 12 ft2 x (x – 1) 7. x + 2 8. 12 ft2 x (x + 4) 9. 24 ft2 (3x – 1) x 10. 15 ft x 2 18 ft2 x (4x + 1)
2nd-Degree-Equation Word Problems B. Given the following area measurements, find x. 2x + 1 11. 5cm2 x 12. 2x – 3 9cm2 x 2x + 1 13. x 18km2 14. (x + 3) 24km2 (5x + 3) 15. 16. 16km2 2 x x + 1 35km2 2 x 2x – 1

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14 2nd degree-equation word problems

  • 2. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree.
  • 3. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt
  • 4. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt where h = height in feet t = time in second v = upward speed in feet per second
  • 5. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second
  • 6. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second?
  • 7. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64,
  • 8. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1)
  • 9. 2nd-Degree-Equation Word Problems Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then h = -16t2 + vt height = -16t2 + vt after t seconds where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 64 ft per second, a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1) h = -16 + 64 = 48 ft
  • 10. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground?
  • 11. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0.
  • 12. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.
  • 13. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t
  • 14. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4)
  • 15. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0
  • 16. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4
  • 17. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds.
  • 18. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained?
  • 19. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point.
  • 20. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.
  • 21. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2)
  • 22. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128
  • 23. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128 = 64 (ft)
  • 24. 2nd-Degree-Equation Word Problems b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t. 0 = -16 t2 + 64t 0 = -16t(t – 4) t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds. c. What is the maximum height obtained? Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h. h = -16(2)2 + 64(2) = - 64 + 128 = 64 (ft) Therefore, the maximum height is 64 feet.
  • 25. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations.
  • 26. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle
  • 27. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle,
  • 28. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW.
  • 29. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW
  • 30. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. If L and W are in a given unit, then A is in unit2. L w A = LW
  • 31. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
  • 32. 2nd-Degree-Equation Word Problems Formulas of area in mathematics also lead to 2nd degree equations. Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2 1 in 1 in 1 in2 L w A = LW
  • 33. Formulas of area in mathematics also lead to 2nd degree equations. If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2 1 in 1 in 1 in2 2 in 3 in 6 in2 2nd-Degree-Equation Word Problems Area of a Rectangle Given a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW. L w A = LW
  • 34. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. L w A = LW
  • 35. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) L w A = LW
  • 36. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 L w A = LW
  • 37. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 L w A = LW
  • 38. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 L w A = LW
  • 39. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 L w A = LW
  • 40. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 L w A = LW
  • 41. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 L w A = LW
  • 42. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. L w A = LW
  • 43. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram L H=height B=base
  • 44. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. L H=height B=base
  • 45. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 w A = LW Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, L H=height B=base
  • 46. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, H=height B=base L w A = LW
  • 47. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we get a rectangle. H=height B=base L w A = LW
  • 48. 2nd-Degree-Equation Word Problems Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width. Let x = width, then the length = (x + 4) LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3 Therefore, the width is 3 and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we get a rectangle. Hence the area A of the parallelogram is A = BH. H=height B=base L w A = LW
  • 49. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. x 2x + 3
  • 50. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27
  • 51. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0
  • 52. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0
  • 53. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
  • 54. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
  • 55. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 Area of a Triangle x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
  • 56. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H
  • 57. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown .
  • 58. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown.
  • 59. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram.
  • 60. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram. If A is the area of the triangle,
  • 61. 2nd-Degree-Equation Word Problems Example C. The area of the parallelogram shown is 27 ft2. Find x. The area is (base)(height) or that x(2x + 3) = 27 x Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. B H 2x + 3 2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft Take another copy and place it above the original one as shown. We obtain a parallelogram. If A is the area of the triangle, then 2A = HB or A = B H . 2
  • 62. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.
  • 63. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.
  • 64. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3)
  • 65. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) x 2x– 3
  • 66. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x x 2x– 3
  • 67. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x x 2x– 3
  • 68. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 2x– 3
  • 69. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) 2x– 3
  • 70. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3
  • 71. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3
  • 72. 2nd-Degree-Equation Word Problems Example D. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height. Let x = height, then the base = (2x – 3) Hence, use the formula 2A = BH 2*10 = (2x – 3) x 20 = 2x2 – 3x 0 = 2x2 – 3x – 20 x 0 = (x – 4)(2x + 5) x = 4 or x = -5/2 2x– 3 Therefore the height is 4 in. and the base is 5 in.
  • 73. 2nd-Degree-Equation Word Problems Exercise A. Use the formula h = –16t2 + vt for the following problems. 1. A stone is thrown upward at a speed of v = 64 ft/sec, how long does it take for it’s height to reach 48 ft? 2. A stone is thrown upward at a speed of v = 64 ft/sec, how long does it take for it’s height to reach 28 ft? 3. A stone is thrown upward at a speed of v = 96 ft/sec, a. how long does it take for its height to reach 80 ft? Draw a picture. b. how long does it take for its height to reach the highest point? c. What is the maximum height it reached? 4. A stone is thrown upward at a speed of v = 128 ft/sec, a. how long does it take for its height to reach 256 ft? Draw a picture. How long does it take for its height to reach the highest point and what is the maximum height it reached?
  • 74. 2nd-Degree-Equation Word Problems B. Given the following area measurements, find x. 5. 8 ft2 x + 2 x 6. 12 ft2 x (x – 1) 7. x + 2 8. 12 ft2 x (x + 4) 9. 24 ft2 (3x – 1) x 10. 15 ft x 2 18 ft2 x (4x + 1)
  • 75. 2nd-Degree-Equation Word Problems B. Given the following area measurements, find x. 2x + 1 11. 5cm2 x 12. 2x – 3 9cm2 x 2x + 1 13. x 18km2 14. (x + 3) 24km2 (5x + 3) 15. 16. 16km2 2 x x + 1 35km2 2 x 2x – 1