Question

Find the percentage error in x when it is estimated to be y and y > x.​

Answer 1

Answer:

VERIFIED ANSWER

Step-by-step explanation:

Let x be the edge of the cubical box, and Δx is error in the value of x.

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01x

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given by

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12x

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as 

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)Δx

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)ΔxHere, dxdS=12x and Δx=0.01x

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)ΔxHere, dxdS=12x and Δx=0.01x⇒  ΔS=(12x)(0.01x)

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)ΔxHere, dxdS=12x and Δx=0.01x⇒  ΔS=(12x)(0.01x)∴  ΔS=0.12x2

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)ΔxHere, dxdS=12x and Δx=0.01x⇒  ΔS=(12x)(0.01x)∴  ΔS=0.12x2The percentage error is,

Let x be the edge of the cubical box, and Δx is error in the value of x.Hence, we have Δx=1001×x∴  Δx=0.01xThe surface area of a cubical box of radius x is given byS=6x2On differentiating A w.r.t x, we get⇒  dxdS=dxd(6x2)⇒  dxdS=6dxd(x2)⇒  dxdS=12xIf y=f(x) is a small increment in x, then the corresponding increment in y, Δy=f(x+Δx)−f(x), is approximately given as Δy=(dxdy)ΔxHere, dxdS=12x and Δx=0.01x⇒  ΔS=(12x)(0.01x)∴  ΔS=0.12x2The percentage error is,Error=6x2

Answer 2

Answer:

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