The dimensions of a rectangular prism are consecutive integers, and its surface area is 2644 units squared.?
Call the consecutive integers n - 1, n , n + 1. V = volume = (n-1)n(n+1) area = 2n(n -1) + 2(n-1) ( n+1) + 2n (n+1) = 2644 2n^2 - 2n + 2n^2 - 2 + 2n^2 + 2n = 2644 6n^2 -2 = 2644 6n^2 = 2646 n^2 = 441 n = 21 so the consecutive integers n - 1, n , n + 1 are 20,21,22 and the volume is (20)(21)(22) = 9240
The length, width and height of a rectangular box are consecutive integers, and the largest dimension is k cm.
Largest is k, so the other 2 must be k-1 and k-2 for them to be consecutive. You didn't mention what the questions was, I'll assume it's: What is the volume of a rectangular box whose sides are consecutive integers, and the largest dimension is k cm. The volume is k*(k-1)*(k-2)=(k^2 - k)*(k-2) =k^3 - 3k^2 + 2*k cm^3
HELP WITH MATHS QUESTIONS!!!!!?
1) The diagonals of a rhombus are perpendicular so if you make a triangle where x is a side of the rhombus which is the hypotenuse of the triangle and the sides are (x+3)/2 and (x+1)/2 you can use the pythagorean theorem to find the length of the side [(x+3)/2]^2 + [(x+1)/2]^2 = x^2 (x^2 + 6x + 9)/4 + (x^2 +2x + 1)/4 = x^2 (2x^2 + 8x + 10)/4 = x^2 (1/2)x^2 + 2x + 5/2 = x^2 -1/2x^2 + 2x + 5/2 = 0 multiply by -2 x^2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5, -1 must be positive for length so side is 5cm 2) 3, 4, 5 that is a a special one no work necessary 3) LW = 224 (L-1)(W+1) = 225 LW - W + L - 1 = 225 LW - W + L = 224 224 - W + L = 224 L = W The original length must have been 16 cm
Math Help please!?
1.) It is quite obvious that the two integers in question are 2 and 3, as 2² + 3² = 13. But here is how to solve this sort of problem in general. Two consecutive positive integers are n and n+1 and n² + (n+1)² = 13 n² + n² + 2n + 1 = 13 2n² + 2n - 12 = 0 ...... divide by 2 n² + n - 6 = 0 ...... ...... factor (n - 2)(n + 3) = 0 => n = 2 or n = - 3 => n should be a POSITIVE integer, so n = 2 and n+1 = 3 ________________ 2.) Let w = the width. Then the length, which 3cm more than twice the width, is 2w + 3 cm. The area = 35cm² width · length = 35cm² w · (2w + 3 ) = 35 2w² + 3w - 35 = 0 (2w - 7)(w + 5) = 0 => w = 7/2 or w = -5 (could also use the quadratic formula instead of factoring - see answer to question 4) The width of a rectangle cannot be negative, so w = 7/2 cm = 3.5 cm And the length is 2w + 3 = 10 cm _____________ 3.) Let b = base. Then the altitude is b+8. The area of a triangle is: Area = 1/2 · base · altitude 24 = 1/2 ·b ·(b+8) .... multiply both sides by 2 to get rid of 1/2 48 = b ·(b+8) 48 = b² + 8b b² + 8b - 48 = 0 (b + 12)(b - 4) = 0 => b = -12 or b = 4 => base of a triangle cannot have neg. length, so b = 4 => the altitude is 4 + 8 = 12 cm ________________ 4.) Let the number be x. Its reciprocal is 1/x. x + 1/x = 3 ..... multiply both sides by x to get rid of the fraction x² + 1 = 3x x² - 3x + 1 = 0 Here we will have to use the quadratic formula. The solutions of a general quadratic equation ax² + bx + c = 0 are x = [ - b ± √(b² - 4·a·c)] / 2·a. For x² - 3x + 1 = 0, a = 1, b = -3 and c = 1, so x = [ - (-3) ± √((-3)² - 4·1·1)] / 2·1 x = [ 3 ± √(9 - 4)] / 2 => x = [ 3 + √5] / 2 and x = [ 3 - √5] / 2 ___________________ Hope this helps.
Right triangles and rectangles helps diagonals? logic?
Hi, The lengths are 3, 4, and 5 This comes from the 3 sides being x, x+1, and x+2. So: x² + (x+1)² = (x+2)² x² + x²+2x+1 = x² + 4x + 4 2x² +2x+1 = x² + 4x + 4 x² -2x - 3 = 0 (x - 3)(x + 1) = 0 x = 3 or x = -1 Since a side can't be negative, discard x = -1. The shortest side is 3 and the other sides are 4 and 5. The perimeter of the rectangle is 28 meters. If a² + b² = c², then 6² + b² = 10² 6² + b² = 10² 36 + b² = 100 b² = 64 b = 8 The perimeter is 2(6)+2(8) = 12 + 16 = 28 meters as the perimteter. I hope that helps!! :-)
Why can't the width, length, and diagonal of a rectangle be consecutive odd integers?
Because of Pythagoras' theorem, OK, here comes the proof. I recommend you understand it... don't just copy it down for your homework. OK. there are three consecutive integers. let the first one equal x. the second one must be x+2, and the third one must be x+4. (if this doesn't make sense, pick any odd integer, and plug it into x. you'll get 3 consecutive odd integers.) Since the diagonal of a rectangle is always the largest number, it must be x+4. therefore by Pythagoras' theorem... (x+4)^2=(x+2)^2 + x^2 x^2+8x+16=2x^2+4x+4 -x^2+4x+12=0 x^2-4x-12=0 so, we get that (x-6)(x+2)=0 x=6,-2 Since x HAS to be positive (imagine a triangle with a negative length... not going to happen) the only numbers that this works for are 6,8,10. Since these are not odd... you get the drift. I hope this helps.