the answer is (3x-2)(9x squared+6x+4)
Your question is ambiguous. (3x)^3 times (3x)^3 = 27x^3 times 27x^3 = 729x^6 or, if you meant 3x^3 times 3x^3 = 9x^6
20x2 - 27x -8 20 * -8 is -160, and -32 and 5 add to -27 20x2 - 32x + 5x - 8 4x(5x-8) + 1(5x-8) (4x + 1)(5x-8)
2x^3-3x^2-11x+7x+32x^2-9x+16 R -41x+3 l 2x^3-3x^2-11x+7-(2x^3+6x^2)-9x^2-11x+7-(-9x^2-27x)16x+7-(16x+48)-412x^2-9x+16-[41/(x+3)]
y will be multiplied by 27. Remember, whatever you do to one side of an equation, you must do to the other. In this case, the relationship between x and y would be expressed as y = x^3 To triple x would look like y = (3x)^3 ...b/c you're tripling the value x alone, not the entire term x^3 y = 27x^3 ...after distributing the exponent (3^3 = 27) In order to set both sides of the equation equal, y must be multiplied by 27... 27y = 27x^3 TADA!
Since the base is equal to the length, then the two parts of the box, up and down are two squares, let's say with length side x. So the other four parts may be rectangles with length x, and wide y, (where y is also the height of the box). Since the surface area is 27 ft^2, we have: Surface Area = 27 = 2x^2 + 4xy Solve for y y = (27 - 2x^2)/4xThus, we have:l = xw = xh = (27 - 2x^2)/4xV = lwhV= (x)(x)[27 - 2x^2)/4x] Simplify and multiply:V = (27x - 2x^3)/4V = (1/4)(27x - 2x^3) Take the derivative: V'= (1/4)(27 - 6x^2) Find the critical values by setting the derivative equal to zero:0 = (1/4)(27 - 6x^2) multiply by 0 both sides:0 = 27 - 6x^2 add 6x^2 to both sides:6x^2 = 27 divide by 6 to both sides:x^2 = 27/6x^2 = 9/2 Square both sides:x = âˆš(9/2)x = (3âˆš2)/2y = (27 - 2x^2)/4xy = 27/4x - (2x^2)/4xy = 27/4x - x/2 substitute (3âˆš2)/2 for x:y = 27/][4(3âˆš2)/2)] - (3âˆš2)/2)/2y = (9âˆš2)/4 - (3âˆš2)/4y = (6âˆš2)/4y = (3âˆš2)/2 As we see the box is a cube.V = side^3V = [(3âˆš2)/2]^3 V = (27âˆš2)/4So, V = (27âˆš2)/4 ft^3 when x = (3âˆš2)/2. Thus, we can maximize the volume of this box if l = w = h = (3âˆš2)/2 ft.
(8x + 9)(3x^2 - 1)
I also need to know the answer to this problem. Can anyone answer it?
3x2 + 27x +60
(2 - 3x)(9x^2 + 6x + 4)
In order to factor the sum of the cubes, we need to use this form a³ + b³ = (a + b)(a² - ab + b²). Let a³ = 27x³ and b³ = 343y³. Then, a = 3x and b = 7y. Perform substitution of these terms for the form, and we obtain: 27x3 + 343y3 = (3x + 7y)(9x2 - 21xy + 49y2)
27x/9x = 3x
(2x - 3)(x - 12)
Step 1: Divide by 9x: 9x (x2 + 3x - 10) = 9x(x + 5)(x - 2)
It is: 23-27x when simplified
(5x - 2)(x - 5)
9x2 + 27x - 36 = 9(x2 + 3x - 4) = 9(x + 4)(x - 1)