The lower bound for the W nearest centroids (cm) algorithm typically refers to the minimum number of comparisons or operations needed to identify the W nearest centroids to a given point in a dataset. This lower bound is influenced by factors such as the dimensionality of the data and the number of centroids. Generally, in a naive approach, the lower bound can be O(N) for each query, where N is the number of centroids, as each point may need to be compared to all centroids. However, more advanced data structures like KD-trees or ball trees can improve this performance in practice.
Suppose the width is W cm Then the length, L = W + 35 cm And the perimeter is 2W + 2L = 2W + 2(W+35) = 4W + 70 4W + 70 = 130 so that 4W = 60 or W = 15 and then L = W + 35 = 50 So the poster is 15 cm by 50 cm
Let the length be ( l ) and the width be ( w ). The perimeter is given by the formula ( P = 2(l + w) ), so ( 20 = 2(l + w) ), which simplifies to ( l + w = 10 ). The area is given by ( A = l \times w = 24 ). Solving these equations, we find the dimensions to be ( l = 12 ) cm and ( w = 2 ) cm (or vice versa).
Let the width of the rectangle be ( w ) cm. Then, the length can be expressed as ( w + 7 ) cm. The area of the rectangle is given by the equation ( w(w + 7) = 60 ). Solving this quadratic equation, we find that the dimensions of the rectangle are a width of 3 cm and a length of 10 cm.
Width = w cmLength = 2w cmArea = (length) x (width) = (2w cm) x (w cm) = 2w2 cm2
To draw a rectangle with an area of 24 cm² and a perimeter of 28 cm, we need to find the dimensions that satisfy both conditions. Let the length be ( l ) and the width be ( w ). The area equation is ( l \times w = 24 ) and the perimeter equation is ( 2(l + w) = 28 ). From the perimeter, we get ( l + w = 14 ). Solving these two equations simultaneously, we can express ( w ) as ( w = 14 - l ) and substitute it into the area equation to find ( l ) and ( w ) are 6 cm and 4 cm, respectively. Thus, the rectangle can be drawn with dimensions 6 cm by 4 cm.
Recall that the perimeter is: P=2*L+2*W If L=27 cm, find W such that P>86 cm P>86 cm 2*L+2*W>86 cm 54 cm+2*W>86 cm 2*W>32 cm W>16 cm So widths greater than 16 cm will make the perimeter greater than 16 cm.
A = w*l so 184 = w*23 or w = 184/23 = 8 cm
Suppose the width is W cm Then the length, L = W + 35 cm And the perimeter is 2W + 2L = 2W + 2(W+35) = 4W + 70 4W + 70 = 130 so that 4W = 60 or W = 15 and then L = W + 35 = 50 So the poster is 15 cm by 50 cm
Let the length be ( l ) and the width be ( w ). The perimeter is given by the formula ( P = 2(l + w) ), so ( 20 = 2(l + w) ), which simplifies to ( l + w = 10 ). The area is given by ( A = l \times w = 24 ). Solving these equations, we find the dimensions to be ( l = 12 ) cm and ( w = 2 ) cm (or vice versa).
W (w + 6) = 952 ie w2 + 6w - 952 = 0 ie (w + 34)(w - 28) = 0 The width is 28 cm and the length 34 cm.
The perimeter of a rectangle is simply the sum of the length of all four sides. So, l + w + l + w, or 2(l + w).
Let the width of the rectangle be ( w ) cm. Then, the length can be expressed as ( w + 7 ) cm. The area of the rectangle is given by the equation ( w(w + 7) = 60 ). Solving this quadratic equation, we find that the dimensions of the rectangle are a width of 3 cm and a length of 10 cm.
Width = w cmLength = 2w cmArea = (length) x (width) = (2w cm) x (w cm) = 2w2 cm2
To draw a rectangle with an area of 24 cm² and a perimeter of 28 cm, we need to find the dimensions that satisfy both conditions. Let the length be ( l ) and the width be ( w ). The area equation is ( l \times w = 24 ) and the perimeter equation is ( 2(l + w) = 28 ). From the perimeter, we get ( l + w = 14 ). Solving these two equations simultaneously, we can express ( w ) as ( w = 14 - l ) and substitute it into the area equation to find ( l ) and ( w ) are 6 cm and 4 cm, respectively. Thus, the rectangle can be drawn with dimensions 6 cm by 4 cm.
The width is 18 cm and the length is 20 cm. The new dimensions are 15 by 24. The formula used is if w=width, w (w + 2) = (w - 3) (w + 6).
Suppose width = W cm Then 4 times width = 4W cm So Length, L = 4W - 3 cm Perimeter = 2*(L+W) = 2*(4W-3 + W) = 2*(5W -3) = 10W - 6 = 114 So 10W = 114+6 = 120 cm and so W = 12 cm.
A tape measure to measure the height (h cm), depth (d cm) and width (w cm). Then capacity = h*d*w cm3 = h*d*w/1000 litres.