Suppose the length and width of the rectangle are L and W metres respectively.Then the perimeter, P = 20 m implies that
2(L + W ) = 20 => L + W = 10 or W = 10 - L.
Then Area = L * W = L * (10 - L) sq metres.
A = (pi)r2 A/pi = r2 √(A/pi) = r
The square is formed from a wire the length of 3x, which forms a perimeter of the same length. Because each side of the square is one fourth of the total perimeter (1 out of 4 equal sides), each side is 3x/4. The square of one side of a square is equal to the area of that square, so the area is (3x/4)2 = (9/16)x2.The answer is then A = (9/16)x2.
sin(2x)=(1/2)sin(x)cos(x), so 6sin(x)cos(x)=12sin(2x)
YES! Simply by taking a quick glance at a graph, you can see several characteristics of the function: local minimums/maximums, points of inflection, end behavior, asymptotes, etc etc... If you wanted to find these without the graph, you would have to do some math which might end up being very time consuming for very complicated functions. Even worse: what if the function is not elementary, and you can't express it in terms of finite arithmetic operations?
If you want to find the lenght of a curve y = f(x) between two values of x, lets say x1 and x2, you must compute this integral : Intx1 to x2[sqrt(dx2 + dy2)] You can either express the original function in terms of y or in terms of x, but it is much simpler to express it in a way such that the integral will not be improper. For example, lets say we want to find the lenght of arc of the curve y = x2 between x = 0 and x = 1. We could express this function in terms of y but we will keep it this way because if we change it, we will have to compute an improper integral, which can sometimes be very tedious. The differential of y = x2 is dy = 2x dx. We now need to square the differential : (dy)2 = (2x dx)2 = 4x2 (dx)2 We now have to compute this integral: Int0 to 1[sqrt(dx2 + dy2)] = Int0 to 1[sqrt(dx2 + 4x2 dx2)] = Int0 to 1[sqrt(1 + 4x2) dx] This last integral is easy to compute using a trigonometric substitution.
It would depend on the shape that you are asking about. Also, only special shapes could express area as a function of the perimeter.Example: a square: area = s2, where s is the length of a side. Perimeter of a square is 4*s.So if P (for perimeter) = 4 * s, then s = P/4,and A (for area) = s2 = (P/4)2 = P2/16But for a rectangle that is not a square, there is no relationship between area and perimeter.
Area of a rectangle is length x width. It isn't clear what the width is in this case - or how you could solve for it.
In a rectangle: P = 2(L + W) and A = LW P = 2(L + W) (replace P with 20) 20 = 2(L + W) (solve for L; divide by 2 to both sides) 10 = L + W (subtract W to both sides ) 10 - W = L A = LW (substitute 10 - W for L) A = (10 - W)W A = 10W - W2
Since the length exceeds the width by 39 inches, if x is the width, then you can express the length as x + 39. Now, use the formula for the perimeter of a rectangle, replacing known values: perimeter = 2(length + width) 234 = 2(x + 39 + x) Solving this for "x" will result in the width; add 39 to get the length.
36 square units. You can't express a perimeter in square units; a perimeter is a length expressed in ordinary units. If the perimeter of this square is 24 units then the answer above is correct.
To express action
To express action
The area ,A, of a rectangle is the length, L, multiplied by the width, W. A=LW hence W=A/L. The perimeter, P, of a rectangle is P= 2(L+W) or 2L+2W. 2W=P-2L and W=(P/2)-L. The Square of the diagonal of a rectangle, D^2 =W^2+L^2 so that W^2=D^2-L^2 and W=sqrt((D^2-L^2). You could also express the width using trigonometric functions, but this is probably enough to get you started.
- tan 60
You use linear units to express it, such as meters or millimeters, if that's what you mean.
It is 1/4 or 0.25
Its height (h) is side (s) times sin 60 orsqrt(3)/2 times side = .866s h = .866s s = h/.866 = 2/sqrt(3) s = 1.154 h perimeter = s + s + s = 3s = 3 x 2/sqrt(3) = 3.46s