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Cutting Gutter Corners

February 15th, 2010 |

Algebra 1 Please Help!?

Equations with this information (There are 2 problems!) 1) A rectangular piece of cardboard is double what is broad. A 4-cm square is cut every corner, and the walls are again making a box. The volume of the box is 616 cm ^ 3. Find the original dimensions of the cardboard. 2) An open rectangular channel is performed by turning up the sides of a piece of metal 20 inches wide. The area of the channel cross section is 50 in. ^ 2 find the depth of the ditch. Please show their work by using equations that factor. Thanks for helping!

1) Carton Length = 2 * Width cardboard After cutting corners height of the box is 4. Also L = length of cardboard – cardboard 8 Length = L + 8 W = Width of cardboard – cardboard 8 Width = W + 8 Connect the board Length and Width of cardboard in the first equation: (L + = 2 (W + L + 8 = 2W + 16 L = 2W + 16 to 8 l = 2W + 8 Volume = W * L * H 616 = W * (2W + * 4 * 616 = 4W (2W + 616 = 8 W ^ 2 + 32W 8W ^ 2 + 32W – 616 = 0 Divide both sides by 8 W ^ 2 + 4 W – 77 = 0 (W + 11) (W – 7) = 0 W Two responses + 11 = 0 W = -11 Answe This is true since the width W can not be negative – 7 = 0 W = 7 L = 2W + 8 L = 2 (7) + 8 L = 14 + 8 L = 22 cartons W = W + 8 cardboard Width = 7 + 8 = 15 Length = Width Cardboard carton length L + 8 = 22 + 8 cardboard Length = 30 So the original board was 15×30 2) L + W + 2L + L = 20 W = 20 W = 20 – 2L cross-sectional area = W * L 50 W * L = 50 = (20 – 2L) * 50 = L 20L – 2L ^ 2 2L ^ 2 – 20L + 50 = 0 L ^ 2 – 10L + 25 = 0 (L – 5) ^ 2 = 0 L – 5 = 0 L = 5 So the depth of the channel is 5 inches I hope this helped Kia

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