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two times a certain integer is 6 more than twice its square. find the integer.

it is kinda confusing.

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Two times a certain integer is 6 more than twice its square ...


Let the unknown integer by n.2n = 6 + 2n2 : 2n2 - 2n + 6 = 0For a quadratic equation an2 + bn + c = 0 then the equation has no real roots ifb2 - 4ac < 0 : In the ...
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Two times a certain integer is 6 more than twice ... - OpenStudy


Two times a certain integer is 6 more than twice its square. Find the integers. I stopped at this equation : 2x^2-2x+6Okay. Let's break down this...
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Twice the square of an integer is five less than eleven times ...


Twice the square of an integer is five less than eleven times the integer The integer is? ... Two times a certain integer is 6 more than twice its square Find the ...
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6 times an integer minus 5 equals the square of the integer ...


6 times an integer minus 5 equals the square of ... Two times a certain integer is 6 more than twice its square Find ... of two times the integer plus two equals Find ...
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The reciprocal of two times a certain integer plus ... - Answers


The reciprocal of two times a certain integer plus the ... I presume that answers ... Two times a certain integer is 6 more than twice its square Find ...
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The reciprocal of two times a certains integer plus the ...


Answers.com ® WikiAnswers ® ... Two times a certain integer is 6 more than twice its square Find the ... Twice the square of an integer is five less than eleven ...
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What integer is 6 more than 0 - answers.com


What integer is 6 more than 0 - answers.com ... 6. ...
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When 11 times a certain integer is subtracted from twice the ...


Answers.com ® WikiAnswers ® ... The reciprocal of two times a certain integer plus the reciprocal of 2 more ... Two times a certain integer is 6 more than twice its ...
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twice the square of a positive number increased by three ...


twice the square of a positive number increased by three times the number ... Twice the Square of a certain whole number ... two more than four times the number.find ...
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Esme is thinking of two integers. One integer is 4 times the ...


if 5 times a certain integer ìs subtracted from twice the ... The square of an integer is 30 more than ... four times the middle integer is two more than the ...
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Suggested Questions And Answer :


10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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two times a certain integer is 6 more than twice its square. find the integer.

2x=2x^2 +6 or 2x^2 -2x+6=0 bekum x^2 -x+3=0 quadratik equashun giv zeroes at 0.5+-1.658312395i   (komplex numbers)
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64239 is divided by a certain number.while dividing the number 175,114 and,213 appears as three successive remainders.find the divisor ?

It's not clear what "successive remainders" means. There is only one remainder after a division: (dividend)=(quotient)*(divisor)+(remainder). For example, if 175 is the remainder after division then 286 divides into 64239 224 times, remainder 175. The largest remainder is 213, implying that the divisor is bigger than 213. It's clear that the divisor cannot be divided again by the same divisor to yield a remainder in excess of the divisor. In the example 286 will not divide into 224, so 224 would be a second remainder. If the divisor has to be used three times to divide successively into each quotient to obtain each remainder then the divisor cannot exceed 40 (the approximate cube root of 64239) and therefore the remainders cannot exceed 40. See explanations in square brackets below. The only other interpretation I can find to the question is that in the process of long division, we have the successive remainders as we proceed with the division. But that doesn't seem to work either. [The numbers a, b and c are integers and b=cx+213; a=bx+114; 64239=ax+175, making ax=64064. Combining these equations we get: a=64064/x=bx+114, so b=64064/x^2-114/x=cx+213, and c=64064/x^3-114/x^2-213/x. Because 64064=ax, we see that 64064 must be divisible by x. 64064 factorises: 2^6*7*11*13. Because the largest remainder is 213, x>213, because x must be bigger than any remainder it creates. But the first term 64064/x^3 can only yield an integer value if x^3<64064, that is, x<40, which contradicts the requirement x>213.] [If the successive remainders appear in the long division, and the divisor has 3 digits as expected, and three remainders suggest that the other factor also has 3 digits, then if 175 is the first remainder it must be the result of dividing the divisor into 642. Therefore the multiple of the divisor =642-175=467. Since 467 is prime then the first digit of the other factor must be 1. 64239 divides by 467 137 times with 260 as the remainder. This is not one of the three remainders, so 175 cannot be the first remainder. Try 114: 642-114=528=2^4*3*11. We need a factor >213. 264, 352. This time we appear to have partial success, because 176 (less than 213, but half of 352) goes into 64239 364 times with a remainder of 175! So we have two of the three required remainders. Unfortunately the third remainder is 87, so 114 cannot be the first remainder. Finally, try 213: 642-213=429=3*11*13. The only factor bigger than 213 is 429 itself, and the final remainder after dividing 64239 by 429 is 318, not included in the three remainders given. So the assumption that the remainders arise during long division is also false.] 64064=2^6*7*11*13. Leaving aside the powers of 2, and just looking at combinations of 7, 11 and 13, we get 7*11=77; 7*13=91; 11*13=143, and 7*11*13=1001. By progressively doubling these products we can discover factors bigger than 213: 308=4*77; 364=4*91; 286=2*143, etc. We can also take 7, 11 and 13 and multiply them by powers of 2: 224=32*7; 352=32*11; 416=32*13. We can write the factors of 64064 in pairs, starting with a factor a little bigger than 213: (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104), (704,91), (1001,64), (1232,52), (2002,32), (2464,26), (4004,16), (4928,13), (8008,8), (16016,4), (32032,2). These are (a,x) or (x,a) pairs. If we take 175 as the remainder, and ignore the other two, then we have a choice of factors. The 3-digit factors are (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104). The only pair in which each factor is greater than 213 is (224,286).
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Twenty percent of a certain number is 16. find the number. And show your work.

Twenty percent of a certain number is 16. find the number. And show your work. 1. Twenty percent of a certain number is 16. Find the number. 2.The length of a rectangle is 5 meters more than twice the width. The perimeter of the rectangle is 46 meters. Find the length and width of the rectangle. 3. One angle of a triangle is 47 degrees of the other two angles, one of them is 3 degress less than three times the other angle. Find the measures of the two angles. Please show all work. 1. 0.2x = 16    5 * 0.2x = 16 * 5    x = 80 2. L = 2W + 5    P = 2L + 2W    46 = 2(2W + 5) + 2W    46 = 4W + 10 + 2W    46 = 6W + 10    36 = 6W    6 = W    L = 2W + 5    L = 2(6) + 5    L = 12 + 5    L = 17    P = 2L + 2W    P = 2(17) + 2(6)    P = 34 + 12    P = 46 3. a = 47    b + c = 180 - 47 = 133    b = 3c - 3    (3c - 3) + c = 133    4c - 3 = 133    4c = 136    c = 34    b = 3c - 3    b = 3(34) - 3    b = 102 - 3    b = 99    a + b + c = 180    47 + 99 + 34 = 180  
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find three consecutive even integers such that twice the largest is equal to two more than three times the smallest.

Let the three integers be 2n-2, 2n and 2n+2. Twice the largest is 4n+4=6n-6+2. 4n+4=6n-4; 2n=8, so n=4 and the numbers are 6, 8 and 10.
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are 46, 48, and 50 the 3 consecutive even numbers that are two times the square root of 24?

me ges yu meen even integers yer werds ruin the quesshun:  square root(24)=4.898979486=reel number not integer so 2*root=9.797958971 ##########46+48+50=144 root(144)=12 2*12=24
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Twenty percent of a certain number is 16. find the number. And show your work.


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Pete challenges his friend Jill to find two consecutive odd integers that have the following relationship.

Two consecutive odd integers can be represented by 2x-1 and 2x+1. Their product is 4x^2-1 and their sum is 4x. So, 4x^2-1=3(4x+6)=12x+18. This is the quadratic: 4x^2-12x-19=0. This has no rational solutions. Or, reading the question slightly differently: 4x^2-1=12x+6, so 4x^2-12x-7=0. In the second case, we can factorise: (2x-7)(2x+1)=0, so x=3.5 or -0.5. The two integers are: 6 and 8, or -2 and 0. None of these are odd, so it isn't possible to find two consecutive odd integers. Why did I think there were two interpretations of the question? "3 times the sum of the integers plus 6" is ambiguous:  is it 3(sum+6) or 3*sum+6? Judging by the result, it was the latter.
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formulate a linear programming problem using graphical method

Let r be the number of minutes of radio and t the number of minutes on TV. r>2t; r<400; 15r+300t<10000 represent the constraints. In terms of effectiveness TV is 25 times more efficient than radio. Using t as the vertical axis and r as the horizontal axis, the lines t=(1/2)r, 15r+300t=10000, r=400 can be plotted. The lines t=(1/2)r and 15r+300t=10000 intersect when r<400. Substituting t=(1/2)r in the second equation we can see that 165r=10000, so r=2000/33, making t=1000/33 at the intersection. The area between the intersection of the two lines, the r axis, and the line r=400 represents where all the constraints are met. Assuming r and t to be in whole minutes, we need to find the integers closest to the intersection point within the defined area. 2000/33=60 and 1000/33=30 so the advertising cost is for 60 mins of radio and 30 mins of TV: Cost=15*60+300*30=900+9000=9900, leaving R100 under budget. But R100 will buy a further 6 minutes of radio advertising so r=66, t=30. 66 minutes is more than twice 30 so the constraint is still valid. 30 minutes of TV advertising has an efficiency equivalent of 25*30=750 radio minutes. Add 66 minutes to this and we have 816 minutes, so the combination is worth 816 minutes of radio time.
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how do you find the area of a prism???

All prisms take a two-dimensional shape and use it as the top and bottom faces in parallel. The top and bottom are then joined to make a solid, so that all the sides are parallelograms. These parallelograms can be rectangles or even squares. The area of the prism is the surface area of the solid, usually including the top and bottom surfaces. The top and bottom faces can each be the same triangle, quadrilateral (including squares, rectangles, trapezoids, etc.), or polygon (star, pentagon, hexagon, etc.). If the top and bottom have n sides, there will be n parallelograms forming the length of the prism. The area of each parallelogram is base times vertical height, where the base length is the length of the particular side of the base figure on which the parallelogram stands. The volume of the prism is the area of the base times the vertical height. For example, if the top and bottom faces are the same regular hexagon (like a pencil), there will be 6 rectangles forming the sides of the prism, each with the same area, A. The surface area of the prism is 6A plus twice the area of the base, or bottom face.
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