During a certain period in the life of an earthworm, its length, L cm, is directly proportional to the square root of N, where N is the number of hours after its birth. One hour after an earthworm is born, its length is 2.5 cm. (i) Find an equation connecting L and N. (ii) Find the length of an earthworm 4 hours after its birth. (iii) How long will it take for an earthworm to grow to a length of 15 cm?
Answers
Let the length be L and breadth be B
Let the length be L and breadth be BGiven, L=8+B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60B=20 cm
Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60B=20 cmSo, Length =20+8=28 cm