My Math Forum Surds Help

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 February 17th, 2011, 07:40 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 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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