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Houston's Leading Hydrographic Decorator

Compound bows started out as 2-wheel designs with an even force draw curve, with approximately 50% allow off with moderate amounts of stored energy. Compound bows have evolved to the point which we see hot innovations every year. The compound bows you sell are state of the art, highly durable, precision created weapons with top quality strings and contents. Proper draw length. There are endless sources that explain how to predict or measure draw length. For the beginning archer, it's smart to visit your local Bass Pro Shops archery department or other qualified archery shop to measure draw

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About Us. Custom Camo is a family owned and operated hydrographics finisher specializing in camouflaged products. Our unique expertise with firearms, atv’s ...
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Looking for the web's Top Customized Matchbooks Sites? Top20Sites.com is the ... Houston's Leading Hydrographic Decorator. ... it's easy to create your own ...
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Houston's Leading Hydrographic Decorator

yu bo in the water ???
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How many seven digit numbers are there when 0, 5, 4 cannot be the leading number?

QUESTION: How many seven digit numbers are there when 0, 5, 4 cannot be the leading number? if a seven-digit number has a leading zero, then it is, in effect, a six-digit number. So ,the question really is: How many seven digit numbers are there when 5 and 4 cannot be the leading digit? if 5 and 4 are excluded as the leading digit then there are 7 other ways of selecting a digit as the leading digit. (there are 10 digits in total but zero is excluded since it is the leading digit that is being considered, hence only 9 left, and only 7 left after 5 and 4 have been removed) There are now 6 digits remaining to be filled in and there are of course 10^6 variations in that selection. Therefore total number of selection is 7*10^6
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John and Ted leave Houston at the same time.

The cosine rule can be used. TJ^2=450^2+800^2-2*450*800cos(305-230), where TJ is the distance between Ted and John. TJ^2=656150.28 approx, so TJ=810km approx.
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how many lead cubes of side 2cm could be made from a lead cube of 8cm?

2 inch kube...get 4 in eech, so number small kubes tu make big kube=4*4*4 =64
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Write this polynomial in standard form and then give the leading coefficient

Standard form is to order the powers of x in descending order: -9x^5-7x^4-7x^3-2x^2+12x+15. The leading coefficient is -9.
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Prove that the number of n-digit binary numbers that have no consecutive 1’s is the Fibonacci number F(n+2).

F3=2 and there are two 1-digit binaries: 0 and 1. So the base case holds. If we call the number of n-digit binary numbers with no consecutive 1's B[n] then B1=2. Assume B[n]=F[n+2], and let's find B[n+1]. F[n]=F[n-1]+F[n-2] by definition. F[n+2]=F[n+1]+F[n]. B[n]=F[n+1]+F[n]=B[n-1]+B[n-2] according to our assumption or hypothesis. Let's look at the set of 3-digit binaries with no consecutive 1's: {000 001 010 100 101}. Now move on to the set of 4-digit binaries by including this set by prefixing a 0 then a 1: {0000 0001 0010 0100 0101 1000 1001 1010 1100 1101}. The set size has doubled but we have caused some consecutive 1's in the second half because of the leading 1. However, if we remove the last two elements we get {0000 0001 0010 0100 0101 1000 1001 1010}. The last 3 elements have 10 consistently as the first pair of digits. The other pair of digits happens to be the set of 2-digit binaries with no consecutive 1's. Therefore the number of 4-digit binaries with no consecutive 1's is B3+B2, so B4=B3+B2. This is of course because we have to place a 0 between the leading 1 and the rest of the binary digits (bits) so that we don't place two 1's together. If we go on to 5-digit binaries the same pattern emerges. 4-digit: {0000 ... 1010}⇒{00000 ... 01010 10000 10010 10100 10101}. B5=B4+B3. Note the second-half leading 10 followed by the 3-digit set with no consecutive 1's. This is true for all natural numbers n. So B[n]=F[n+1]+F[n]=B[n-1]+B[n-2]. And B[n]=F[n+2].    
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construct a 3rd degree polynomial with zeros of 4-i, 4+i, 2 and a leading coefficent of -4

There are three zeroes and it's a third degree polynomial so there are no more zeroes. The complex zeroes can be multiplied together, cancelling out the complex terms: (x-4+i)(x-4-i)=(x-4)^2+1=x^2-8x+17. Now multiply by the final factor (x-2) to give x^3-8x^2+17x-2x^2+16x-34=x^3-10x^2+33x-34. The leading coefficient is -4 so we need to multiply through by this number: -4x^3+40x^2-132x+136.
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how many feet is the blue bat leading the red boat in 3 1/2 hours.

A red boat and blue boat are in a race. After 10 minutes, the blue boat is 500 feet ahead of the red boat. If their speeds remain constant how many feet is the blue boat leading the red boat after 3 1/2 hours? We need to convert 3 1/2 hours into a comparable number of 10 minute intervals. 3 1/2 hours = 210 minutes 210 minutes / 10 minutes = 21 Because the blue boat gains 500 feet every 10 minutes, we multiply 500 feet by the number of 10 minute intervals. 500 ft * 21 = 10500 ft
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consider a polynomial with integer coefficients in which the leading corfficient is not equal to 1. Can the related polynomial equation have a rational root that is an integer?

Yes. For example, 2x^3+2x^2-5x-5=0 factorises into (x+1)(2x^2-5)=0. There is one rational root, which is an integer, (x=-1) and 2 irrational roots.
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show that this information leads to the inequality x+4y<=180.

????????????????? wot info ???????????????
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