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deff fn [24]
3 years ago
8

The side of a square is measured to be 10 ft with a possible error of ±0.1 ft. Use linear approximation or differentials to esti

mate the error in the calculated area. Include units in your answer
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer:

The error in the calculation of the area will be 2 square feet.

Step-by-step explanation:

Let the actual length of the square is L ft.  

So, actual area A = L² ......... (1)

Now, if there is an error in measuring length is ΔL, then  

(A + ΔA) = (L + ΔL)² {Since if there is an error in length by ΔL, then there will be an error in the calculation of area by ΔA}

⇒ A + ΔA = L² + 2×L×ΔL + (ΔL)²

Now, by linear approximation, neglect the term (ΔL)² as it will be very small.

So, A + ΔA = A + 2×L×ΔL {Since A = L²}

⇒ ΔA = 2×L×ΔL .......... (2)

Now, given that  L = 10 ft and ΔL = 0.1 ft.

Hence, ΔA = 2× 10× 0.1 = 2 sq. ft.

Therefore, the error in the calculation of the area will be 2 square feet. (Answer)

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Find the distance between two points ( -1,1) and (2,-4)
sladkih [1.3K]

Answer:

The distance between two points ( -1,1) and (2,-4) is:

d=\sqrt{34}  or d = 5.8 units.

Step-by-step explanation:

Given the points

  • (-1, 1)
  • (2, -4)

Finding the distance between (-1, 1) and (2, -4) using the formula

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

substitute (x₁, y₁) =  (-1, 1)   and  (x₂, y₂) = (2, -4)

   =\sqrt{\left(2-\left(-1\right)\right)^2+\left(-4-1\right)^2}

   =\sqrt{\left(2+1\right)^2+\left(-4-1\right)^2}

   =\sqrt{3^2+5^2}

   =\sqrt{9+25}

   =\sqrt{34}  units

or

d = 5.8 units

Therefore, the distance between two points ( -1,1) and (2,-4) is:

d=\sqrt{34}  or d = 5.8 units.

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