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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.
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Answer this a piece of poster has a perimeter of 130 cm what are the dimensions of the poster if its length is 35 cm longer than its width?
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
Perimeter of 20 cm area 24 cm what is the length and width in 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).
What is the dimension of the rectangle when the length is 7 cm more than its width and the area is 60cm squared?
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.
What is the area of a rectangular desk top that is w cm wide and twice as long as it is wide?
Width = w cmLength = 2w cmArea = (length) x (width) = (2w cm) x (w cm) = 2w2 cm2
How do you draw a rectangle with area 24 cm2 and perimeter 28 cm?
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 length of a rectangle is fixed at 27 cm what widths will make the perimeter greater than 86cm?
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.
What is the value of w if A equals 184 cm2 and l equals 23 cm?
A = w*l so 184 = w*23 or w = 184/23 = 8 cm
Answer this a piece of poster has a perimeter of 130 cm what are the dimensions of the poster if its length is 35 cm longer than its width?
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
Perimeter of 20 cm area 24 cm what is the length and width in 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).
The length of a rectangle is 6 cm greater than its width its area is 952 cm squared find its dimension?
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.
Find the perimeter of a rectangle 12.5 cm long and 3.5 cm wide?
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).
What is the dimension of the rectangle when the length is 7 cm more than its width and the area is 60cm squared?
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.
What is the area of a rectangular desk top that is w cm wide and twice as long as it is wide?
Width = w cmLength = 2w cmArea = (length) x (width) = (2w cm) x (w cm) = 2w2 cm2
How do you draw a rectangle with area 24 cm2 and perimeter 28 cm?
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 length of the rectangle is greater than its breadth by 2 cm if the length is increased by 4 cm and the breadth decreased by 3cm the area remain same find the length and breadth of the rectangle?
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).
The length of a rectangle is 3 cm less than 4 times the width The perimeter is 114 cm Find the width?
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.
What do you use to measure the capacity of a school locker?
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.
