6. The dimensions of a box are 3 cm x 5 cm x 7 cm. The...

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6. The dimensions of a box are 3 cm x 5 cm x 7 cm. The width, length, and height of the box must be increased by the same amount x in order for the box to have a volume of 693 cm cubed. Determine the value of x that will produce a box with a volume of 693 cm³. [T4] [C2] (3+xX5+x)(7+x)-8693 X²+15x+71x+105-693 X+15x+71x-588-0 (x-4)(x+19x+147)=0 X-40 or x²+19x+1470 X=4 So the value is 4 CHECK IF THERE ARE OTHER VALUES!!! 7. The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436 m using the algebraic method. [T6] [CZ] We need the least volume of 8436m3 Let the width be x length becomes 1+3x Height becomes 2x-5 Volume-length x widthx height V= (3x+1)(x)(2x-5) We are given that volume-8436 (3x+1)(x)(2x-5)-8436 X(3x+1)(2x-5)-8436 X(6x³-13x-5)-8436 6x³-13x³-5x-8436-0 The polynomial 6x²-13x³-5x-8436 can be written as (x-12)(6x2+59x+703)>0 For this is product to be >0 Both (x-12)and(6x +59x+703) Should be greater than zero (x-12)>0-12 CHECK IF THERE IS ANY VALUES FOR X Hence any width>12 would serve the purposed. (-3) (-3) 6. The dimensions of a box are 3 cm x 5 cm x 7 cm. The width, length, and height of the box must be increased by the same amount x in order for the box to have a volume of 693 cm cubed. Determine the value of x that will produce a box with a volume of 693 cm³. [T4] [C2] (3+xX5+x)(7+x)-8693 X²+15x+71x+105-693 X+15x+71x-588-0 (x-4)(x+19x+147)=0 X-40 or x²+19x+1470 X=4 So the value is 4 CHECK IF THERE ARE OTHER VALUES!!! 7. The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436 m using the algebraic method. [T6] [CZ] We need the least volume of 8436m3 Let the width be x length becomes 1+3x Height becomes 2x-5 Volume-length x widthx height V= (3x+1)(x)(2x-5) We are given that volume-8436 (3x+1)(x)(2x-5)-8436 X(3x+1)(2x-5)-8436 X(6x³-13x-5)-8436 6x³-13x³-5x-8436-0 The polynomial 6x²-13x³-5x-8436 can be written as (x-12)(6x2+59x+703)>0 For this is product to be >0 Both (x-12)and(6x +59x+703) Should be greater than zero (x-12)>0-12 CHECK IF THERE IS ANY VALUES FOR X Hence any width>12 would serve the purposed. (-3) (-3)

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7 24 let increase amount x length Sx un Width 3 x Com Height V lwk 693 5x 3x 7H 693 xx 15 8 x 7 ... View the full answer