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 February 17th, 2011, 12:45 AM #1 Newbie   Joined: Feb 2011 Posts: 1 Thanks: 0 Surds Help I really need help for this and this is my first time on this website, so sorry if it's in the wrong category. Sorry in advance for some of the questions; they were typed off the sheet. And if you could show your working out, that would be great. Q1. So that any digit from 0 to 9 can be displayed on a scoreboard, tube lighting is set up on a digital clock style The digits are that of a digital clock, so 8= 7tubes, 4=4, 5=5 and 0=6 If each individual section of tubing is (3?2-2) metres in length, what exact length of tubing would be lit to produce the following digits? (Make sure in each case that you expand out any brackets in your answer and keep it in surd form) Q2. Find the exact area of the following shape. (Make sure you expand any brackets in your answer and keep it in surd form) Top length. (?5 + 6)cm and the side length is 2?7cm. Q3. Find the exact length x of the rectangle. Its exact area is equal to 14?60 + 7?20 cm²/ (Make sure you express your answer in simplest form) The side length is 7?5cm. Q4. Which number is bigger, 7?2 or 3?11. Show all your reasoning. Q5.A cube has a sidelength equal to (8?5 + 9?11)cm a) Find the area of one of the faces of the cube. Expand any brackets and give your answer as an exact value. (keeping it in surd form) b) Hence, find the total surface area of the cube as an exact value. c) Find, correct to one decimal place, the area of wrapping paper that would be required to wrap up this cube as a present. I really need help for this and answers would be great. Thanks in advance.
 February 17th, 2011, 08:40 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Surds Help Q1. For this, you take the number of sections required to display a given digit and multiply it by the length of an individual section. If n is the number of sections it takes to display the digit, then the total length of tubing would be: $n$$3\sqrt{2}-2$$=3n\sqrt{2}-2n$ Q2. Assuming the area is rectangular, then the area A in centimeters squared would be: $A=$$\sqrt{5}+6$$$$2\sqrt{7}$$=2\sqrt{35}+12\sqrt{7 }$ Q3. Here we would use: $x=\frac{A}{h}=\frac{$$14\sqrt{60}+7\sqrt{20}$$\tex t{ cm^2}}{7\sqrt{5}\text{ cm}}=\frac{7\sqrt{5}$$2\sqrt{12}+\sqrt{4}$$}{7\sqr t{5}}\:\text{cm}=$$4\sqrt{3}+2$$\text{ cm}$ Q4. For two real numbers, a and b, where 1 < a,b then if a > b then a² > b². So, squaring both numbers given, we have: $$$7\sqrt{2}$$^2=7^2\cdot2=98$ $$$3\sqrt{11}$$^2=3^2\cdot11=99$ So we conclude $7\sqrt{2}<3\sqrt{11}$ Q5. a) $A=$$\(8\sqrt{5}+9\sqrt{11}$$\text{ cm}\)^2=$$8^2\cdot5+2\cdot8\cdot9\sqrt{55}+9^2\cdo t11$$\text{ cm^2}=$$1211+144\sqrt{55}$$\text{ cm^2}$ b) $6A=6$$1211+144\sqrt{55}$$\text{ cm^2}=$$7266+864\sqrt{55}$$\text{ cm^2}$ c) The minimum amount of wrapping paper it would take me to wrap this cube, assuming I'm not going to paste a square of paper to each face of the cube, but rather I'm going to wrap it in a more traditional manner, would be: $A=$$2\(8\sqrt{5}+9\sqrt{11}$$\text{ cm}\)$$4\(8\sqrt{5}+9\sqrt{11}$$\text{ cm}\)=8A=$$9688+1152\sqrt{55}$$\text{ cm^2}\approx18231.5\text{ cm^2}$

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