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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 ...

### 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...

### Twice the square of an integer is five less than eleven times ...

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### 6 times an integer minus 5 equals the square of the integer ...

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### What integer is 6 more than 0 - answers.com

What integer is 6 more than 0 - answers.com ... 6. ...

### When 11 times a certain integer is subtracted from twice the ...

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### twice the square of a positive number increased by three ...

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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)

### 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).

### 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

### 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.

### 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

### 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.

### 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.

### 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.