100 V2 - 25 W2 = (10V + 5W) (10V - 5W)Another way to show it:100 V2 - 25 W2 = 25 (4 V2 - W2) = 25 (2V - W) (2V + W)
w2/(3w)2 = w2/9w2 = w2-2/9 = w0/9 = 1/9 or w2/3w2 = 1/3
(w + 9)(w + 9)
w x w + 6 = 135 w2 + 6w - 135 =0 Ie (w - 9)(w + 15) = 0 width = 9m (length = 15m)
Six less than a number w squared is written algebraically as w2 - 6
If by w2 you mean "w squared", then it cannot be done. BECAUSE w2 + 49 = 0 w2 +49 - 49 = 0 - 49 w2 = -49 w2 means a number multiplied by itself. Any number multplied by itself will result in a POSITIVE number. positive x positive = positive negative x negative = negatve
No. The Social Security Administration will send you a 1099-SSA or 1099-RRB for use in completing your taxes.
When trying to finalize or finish your W2 form and to get help, the IRS is usually best place to start by going to http://wwww.irs.gov Other than using the irs you can also go to w2express.com
100 V2 - 25 W2 = (10V + 5W) (10V - 5W)Another way to show it:100 V2 - 25 W2 = 25 (4 V2 - W2) = 25 (2V - W) (2V + W)
It is not possible to solve for a value of X since there are three unknowns, X, v and W and only one equation. All that can be done is to make x the subject of the equation - that is, to express X in terms of the other two variables. X2 - v = W2 X2 = W2 + v X = +/- sqrt(W2 + v)
It does not matter if you file with your original W2 or not. As long as you have a copy of the W2, that is all that is needed. If you file online, you do not even need to send your W2 in.
There are websites that will walk you through completing a 1040 tax form. You will need your W2 and 1099 forms, in addition to any interest income and all related tax documents.
The length and width of the rectangle (and thus it's area) will be defined by the circle. Specifically, the diagonal of that rectangle will always be equal to the diameter of the circle. To find the maximum area then, one can simply define the rectangle with that relationship, take the derivative of that function, and find out where that comes to a value of zero. The area will be peaked at that point: a = lw d2 = l2 + w2 ∴ l = (d2 - w2)1/2 Remember that the radius is 12, so the diameter is 24, and things can be simplified by plugging that into the equation at this point: l = (576 - w2)1/2 ∴ a = (576 - w2)1/2 w The next step is to take this equation for area, and find it's rate of change with respect to width: ∴ da/dw = (576 - w2)1/2 + 1/2 * (576 - w2)-1/2 * -2w * w ∴ da/dw = (576 - w2)1/2 - (576 - w2)-1/2 * w2 Now let that value equal zero: 0 = (576 - w2)1/2 - (576 - w2)-1/2 * w2 ∴ (576 - w2)1/2 = (576 - w2)-1/2 * w2 ∴ 576 - w2 = w2 ∴ w2 = 576 / 2 ∴ w2 = 288 ∴ w ≈ 16.97 And one can work out the corresponding height: d2 = l2 + w2 ∴ 242 = l2 + 288 ∴ l2 = 576 - 288 ∴ l2 = 288 ∴ l ≈ 16.97 Meaning that the optimum rectangle is a perfect square. To be completely thorough, it should be confirmed that the area found was a maximum and not a minimum. This can be done easily enough by taking a slightly lesser width and a slightly greater width, and seeing how their areas compare: a = (576 - w2)1/2 w Let w = 16 ∴ a = (576 - (16)2)1/2 * 16 ∴ a = 3201/2 * 16 ∴ a = 51/2 * 128 ∴ a ≈ 286.22 Let w = 18 a = (576 - 182)1/2 * 18 ∴ a = 2521/2 * 18 ∴ a = 71/2 * 108 ∴ a ≈ 285.74 Both of these values are less than the area surrounded by the square, so the square's area is indeed a maximum and not a minimum.
where do I replace ss w2
The square root of 150 will be between 12 and 13, closer to 12.
w2 form
what day when w2 form