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WRITE AN EXPRESSION THAT MEANS -1.25/1 TIMES THE RACE NUMBER MINUS 3.5

Please help:   SOLVE : WRITE AN EXPRESSION THAT MEANS -1.25/1 TIMES THE RACE NUMBER MINUS 3.5

Research, Knowledge and Information :


Writing Expressions - BIG IDEAS MATH


x + 14 The phrase “more than” means addition. b. A number y minus 75 ... 6 times 50 3. 25 less than a number b ... Section 1.2 Writing Expressions 13 Write the ...
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Writing Expressions (solutions, examples, videos)


Writing Expressions. ... Expression . 6 more than 5 times a number . 5x + 6 . ... 1: Half of a number is 16. Write an equation to represent the situation.
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Writing Algebraic Expressions


Writing Algebraic Expressions is presnted by Math Goodies. Learn how to translate verbal phrases into algebraic expressions. ... three times a number decreased by ...
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Algebra - Math League


To evaluate an expression at some number means we replace a ... second expression for "four times my age, minus ... the list is the number 1. 1, 2, 3, 4, 5, 6 ...
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Numerical Expression - Math Dictionary - ICoachMath.com


... what is an numerical expression ... means 'phrase'. A numerical expression is a ... number. For example, the numerical expression ...
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Algebraic expressions - A complete course in algebra


... x plus 5 times x minus 2? (x + 5) ... To "evaluate" means to name and write a number. Example 1. ... 3x + 1 . Now an algebraic expression is not a sentence, ...
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Writing Expressions and Equations


Lesson 7.1 Writing Expressions and Equations 319 Write the verbal phrase ... 15. 13 is equal to 5 minus a number. 16. 1 4 ... Lesson 7.1 Writing Expressions and ...
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Name: Unit3 #2 Writing Expressions and Equations


Writing Expressions and Equations ... The ratio of z and 3 Practice: Underline key words. Write each phrase as an algebraic ... is 5 times a product of 5 and a number
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Translating Words into Mathematical Symbols


Translating Words into Mathematical Symbols ... In the expression 2x + 3 ... “To factor” means to write a number as the product of its factors.
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Suggested Questions And Answer :


how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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WRITE AN EXPRESSION THAT MEANS -1.25/1 TIMES THE RACE NUMBER MINUS 3.5

maebee yu meen -1.25*(x-3.5) ??????
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how to solve this quadratic equation x^2-2^x-13=0

I think you may have typed this in wrongly because the term 2^x is not a permitted term in a quadratic equation. As it stands the solution to the equation is about -3.6168. Let's say you made a mistake in typing and the middle term is 2*x or 2x, then the solution exists but it is irrational (involves square roots). The easiest way to solve x^2-2x-13=0 is to complete the square: (x^2-2x+1)-1-13=0, which is (x-1)^2-14=0, so (x-1)^2=14. Take square roots of each side: x-1=+sqrt(14)=+3.7417 approx. A square root always has two solutions, one positive and one negative, but the same magnitude number. So x has two possible values: x=1+3.7417=4.7417 or x=1-3.7417=-2.7417. If you meant the question to be x^2-12x-13=0, then the solution is (x-13)(x+1)=0 and x=13 or -1. To work this out we ask: what are the factors of 1 (x^2 coefficient) and 13 (the constant term). The factors of 1 are 1 and 1 because only 1 times 1 make 1; and the factors of 13 are only 1 and 13. The coefficient of the x^2 term tells is how many x's go in each bracket. That's 1x or just x in each bracket. And the factors of 13 tell us what. Numbers to write in each bracket, so that's 1 and 13. So we have (x 1)(x 13). What about the signs between them? We look at the sign in front of 13 in the quadratic. It's minus, and that means there will be a plus in one bracket and a minus in the other. But which way round? Well, there's one more test: we take the factors of 13 and subtract them because the minus sign in front of 13 tells us we need to subtract. If it had been plus, we would have added the factors. 13-1=12. If 12 is the coefficient of the x term then the quadratic can be solved. (If the number had not been 12 we could not have solved the quadratic this way.) The sign in front of 12x is the sign that goes in front of the larger number in the brackets, so minus goes in front of 13. So we have (x+1)(x-13)=0. One or other of these factors is zero, so x+1 or x-13 is zero. x+1=0 means x=-1 and x-13=0 means x=13. These are the solutions or roots.
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write a system of two linear equations showing the distance of each animal can travel to model a fair race.

One way to resolve this is to work out the distance each must run to take the same time to complete the race. Distance=speed times time, so time=distance÷speed. So if the distances are D1 and D2 we can write D1/43=D2/35 (time for coyote=time for rabbit) and that means D1/D2=43/35 (about 1.23). If the ratio of the distances is the inverse of the speeds then the race will be fair. (D1=43T (coyote) and D2=35T (rabbit); divide one equation by the other: D1/D2=43T/35T so D1/D2=43/35.) TRACK ARRANGEMENTS FOR FAIR RACE We could have two parallel linear tracks but the rabbit starts at a different point along its track. If the longer track is 43n miles long then the shorter track is 35n miles long where n is a number, probably a fraction, for example 1/10. The rabbit's start-line is 43n-35n=8n miles beyond the coyote's. The rabbit runs only 35n miles while the coyote runs 43n miles. The finishing-lines are in line. The race will take n hours if both animals run at top speed on average throughout the race. So if, for example, n=1/10 the race will last for 1/10 hr = 6 minutes=4.3/43=3.5/35. The rabbit runs 3.5 miles while the coyote runs 4.3 miles. Another way is to use circular tracks. Since the circumference of a circle is a fixed multiple (2π) of the radius, we have two concentric circular tracks where the radius of the inner track is R1 (for the rabbit) and radius of the outer track is R2 so that R1/R2=35/43=2πR1/2πR2, where 2πR1 and 2πR2 are the lengths of the inner and outer tracks. The advantage of circular tracks is that the race can be run over a number of laps rather than just one lap. The race is more exciting because spectators can surround the track and observe the progress of the race better. A third solution is to have semicircular tracks placed at the two ends of a rectangular circuit. The curvature of the circles compensates for the difference in speeds of the animals. The width of the rectangle is 2R1, where R1 is the radius of the inner track. If L=length of the rectangle then the length of the tracks are (2L+2πR1) and (2L+2πR2), and (2L+2πR1)/(2L+2πR2)=35/43=(L+πR1)/(L+πR2), so 43L+43πR1=35L+35πR2; 8L=π(35R2-43R1) (35R2>43R1, so R1/R2<35/43). This equation relates the length of the rectangular part of the track to the inner and outer radii of the semicircular parts of the tracks. If this equation is satisfied the start and finish lines will be in line. If the equation is not satisfied then, the start and finish lines will be out of line by the value of the expression 8L-π(35R2-43R1). (EXAMPLE: If R2=2R1 for the above type of track, 8L=π(70R1-43R1)=27πR1, and L=27πR1/8=10.6R1 approx. or 5.3W where W=2R1 is the width of the rectangle.) This answer probably goes into more detail than you require, but I hope you get the general idea.  
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please help me

Question: find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4 ? Let us rewrite your problem statement. find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4, or v1, v2, v3, v4, v5 Turning those statements above into mathematical expressions, 3 times the product of the 3rd and 4th,                            becomes   3*v3*v4 minus the product of the 2nd and (4 less than the first),  becomes   - v2*(v1-4) plus 3 times the 5th                                                          becomes   + 3*v5 is 232.                                                                               becomes   = 232 As a single expression, this is 3*v3*v4 - v2*(v1 - 4) + 3*v5 = 232 3(x+2)(x+3) - (x+1)(x - 4) + 3(x+4) = 232 3(x^2 + 5x + 6) - (x^2 - 3x - 4) + (3x + 12) = 232 2x^2 + 21x + 34 = 232 2x^2 + 21x - 198 = 0 (2x + 33)(x - 6) = 0 x = -33/2, x = 6 But x is an integer, so the only solution is x = 6 The numbers then are : 6, 7, 8, 9, 10
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Creata a fraction story

FIVE LITTLE PIGS GO TO THE MALL Mama Pig gave her five little pigs seven and a half dollars between them to spend at the mall. It was a cold day, twenty-three Fahrenheit, minus five Celsius, or five degrees below freezing. Off they trotted at a quarter to three in the afternoon. "How far is it?" the youngest pig asked after a while. "One point seven five miles from home," said the eldest. "What does that mean?" asked the youngest. "Well," explained the eldest, "if we divide the distance into quarter miles, it's seven quarters." "How long will it take to get there?" asked the pig in the middle. Her twin sister replied, "It's five past three now, so that means we've taken twenty minutes to get here. Remember the milestone outside our house? There's another one here, so we've come just one mile and we've taken a third of an hour. [That means our speed must be three miles an hour.] "How much longer?" the youngest asked. "Three quarters of a mile to go," the next youngest started. "Yes," said the eldest, "we get the time by dividing distance by speed, so that means three quarters divided by three, which is one quarter of an hour [which is fifteen minutes]." ["So what time will we arrive?" the youngest asked. "About twenty past three," all the other pigs replied together.] "That's thirty-five minutes altogether," the eldest continued, "which means that - let me see - seven quarters divided by three is seven twelfths of an hour. One twelfth of an hour is five minutes [so seven twelfths is thirty-five minutes]. Yes, that's right." When they got to the mall, it had started to snow. Outside there was a big thermometer and a sign: "COME ON IN. IT'S WARMER INSIDE!" The eldest observed: "It shows temperature in Fahrenheit and Celsius. See, there's a scale on each side of the gauge. It's warmer now than it was when we left home. The scales are divided into tenths of a degree. It says twenty-seven Fahrenheit exactly, and, look, that's the same as minus two point eight Celsius," speaking directly to the youngest, "because the top of the liquid is about eight divisions between minus two and minus three. Our outside thermometer at home is digital, but this is analogue." The youngest stuttered: "What's 'digital' and 'analogue'?" "Well," began one of the twins, "your watch is analogue, because it has fingers that move round the watch face. Our thermometer at home is digital, because it just shows the temperature in numbers." ["Yes," said the eldest, "it would show thirty-seven point zero degrees Fahrenheit, four degrees warmer, and minus two point eight Celsius, two point two degrees warmer than when we left."] Continued in comment... 
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how do you solve 3t^4+2t^3-300t-50=0

The expression does not factorise and the roots seem to be t=-0.16669 and 4.48653 approx. Check that you haven't missed out a term (e.g., t^2). If the equation was meant to factorise completely, then the product of the factors=-50. The factors must therefore consist of 1, 2, 5, 5. The factors of the 3t^4 must consist of 3t, t, t, t. The minus sign in front of the constant 50 implies that there must be an odd number of pluses and minuses amongst the factors: one minus and 3 pluses, or three minuses and one plus.  Let the coefficient of the apparently missing t^2 be k. The equation then reads: 3t^4+2t^3+kt^2-300t-50=0. If (t-1) is a factor then 3+2+k-300-50=0 and k=345. If (t+1) is a factor then 3-2+k+300-50=0 and k=52-303=-251. If (t-2) is a factor then 48+16+4k-600-50 and k=(1/4)(650-64)=293/2. If (t+2) is a factor then 48-16+4k+600-50 and k=(1/4)(66-648)=-291/2. It seems improbable that (t-2) or (t+2) could be a factor because k would be a mixed number (or improper fraction), so k=345 or -251. If (t-5) is a factor then 1875+250+25k-1500-50=0 and k=(1550-2125)/25=-23. If (t+5) is a factor then 1875-250+25k+1500-50=0 and k=(300-3375)/25=-123. We know that (3t+f) must be included as a factor, but if f=-5 or +5 then, since we need two 5's within the set of factors, so if one contains 3t, the other can only contain t. That means that (t-5) or (t+5) must be a factor. There's no point trying other factors like (t-25) or (t+10), because they would require (t-2) or (t+2), or (t-5) or (t+5), and (t-1) and (t+1) as other factors, which have already been covered. The exercise shows that no one value of k allows us to find more than one root. Had any two values of k above been the same we would have had at least two roots and a quadratic, which could have been solved, if it had had real roots. The conclusion is that finding one root only leaves us with a cubic equation that has no rational, real roots. This means that the question has been wrongly stated, and the inclusion of kt^2 doesn't help to solve it satisfactorily.
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solve by factorisation the following equation (1).5x^2+2x-15=0 (2).21x^2-25x=4 (3).10x^2+3x-4=0

  (1) 5x^2+2x-15=0 I suspect that this should be 5x^2+22x-15=0, because the equation as written does not factorise. Write down the factors of the squared term coefficient and the constant term. We write these as ordered pairs (a,b) squared term (c,d): (1,5) and constant: (1,15), (15,1), (3,5), (5,3). The sign of the constant (minus) tells us that we are going to subtract the cross-products of  factors. If it had been plus we would be adding the cross-products. Now we create a little table: Quadratic factor table a b c d ad bc  |ad-bc| 1 5 15 1 1 75 74 1 5 1 15 15 5 10 1 5 3 5 5 15 10 1 5 5 3 3 25 22 The column |ad-bc| just means the difference between ad and bc, regardless of it being positive or negative, just take the smaller product away from the larger. If the coefficient of the x term is in the last column, then that row contains the factors you need. If it isn't in the last column, then the quadratic doesn't factorise, or you've missed some factors. If 22x is the middle term, then the factors are shown in the last row and (a,b,c,d)=(1,5,5,3). Now we look at the sign of the middle term. Whatever the sign is we put it in front of the number c or d for the larger cross-product. The cross-products are bc and ad. In this case, bc is bigger than ad so the + sign goes in front of c. The sign in front of the constant tells us whether the sign in front of d is different or the same. If the sign is plus it's the same, otherwise it's the opposite sign. So in this case, it's minus, so the minus sign goes in front of d and we have (ax+c)(bx-d) (note the order of the letters!) or (x+5)(5x-3), putting in the values for a, b, c and d. If (x+5)(5x-3)=0, then x+5=0 or 5x-3=0. So in the first case x=-5 and in the second case 5x=3 so x=3/5. (2) 21x^2-25x-4=0 (moving 4 over to the left to put the equation into standard form). Quadratic factor table a b c d ad bc  |ad-bc| 1 21 1 4 4 21 17 1 21 4 1 1 84 83 1 21 2 2 2 42 40 3 7 1 4 12 7 5 3 7 4 1 3 28 25 3 7 2 2 6 14 8 The row in bold print applies. The sign in front of 4 is minus so the signs in the brackets will be different. 28 is the larger product, so since we have -25x, the minus sign goes in front of c (4) and plus in front of d (1): (3x-4)(7x+1)=0. So the solution is x=4/3 or -1/7. (3) 10x^2+3x-4=0. Quadratic factor table a b c d ad bc  |ad-bc| 1 10 1 4 4 10 6 1 10 4 1 1 40 39 1 10 2 2 2 20 18 2 5 1 4 8 5 3 2 5 4 1 2 20 18 2 5 2 2 4 10 6 (2x-1)(5x+4)=0, so x=1/2 or -4/5.  
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k^2-k-20/18

I read this as (k^2-k-20)/18. What pair of factors of 20 differ by 1? This is essentially what the quadratic in k is asking. The pairs of factors of 20 are (1,20), (2,10), (4,5) and the only pair with a difference of 1 is (4,5). The larger of the two has a minus and the other a plus to make the difference -1 which is the coefficient of the middle term. So k^2-k-20 factorises: (k-5)(k+4). And we place this over 18: (k-5)(k+4)/18. This won't factorise any further. When I was reviewing this question I felt there was more to it. Supposing the question was: what values can k take so that the expression is an integer? I spotted something interesting. The zeroes of the quadratic are 5 and -4: either of these values make the expression zero. But 5+4=9, and 9 is a factor of 18, the denominator. We can write the factors of the quadratic as y(y-9)/18, where y replaces k+4; or we can write y(y+9)/18, where y replaces k-5. The pairs of factors of 18 are (1,18), (2,9), (3,6). If the numerator contains any of these pairs, the expression will be an integer, or whole number, positive or negative. What values of y must we have so that y(y-9)/18 is an integer? Let's start with y=0; the expression is zero, which is an integer. Now y=3; we have 3(-6)/18=-1, another integer. y=6; 6(-3)/18=-1. y=9; we get 0. y=12 we get 12*3/18=2. y=15; we get 15*6/18=5. In fact, all multiples of 3 work, so we can write y=3N as the general solution, where N is a positive or negative integer. What about k? We know that y=k+4 or k-5 so k+4=3N or k-5=3N are solutions. We can write these solutions as k=3N-4 or k=3N+5. If N=0 we have -4 and 5, which are the zeroes of the quadratic. If N=1, k=-1 or 8; if N=2, k=2 or 11; if N=3, k=5 or 14; if N=4, k=8 or 17; and so on. If you substitute these and other values of k (according to the formula) you'll see that the expression is always an integer.
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What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.      
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