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DerKrebs [107]
3 years ago
9

Use a linear approximation (or differentials) to estimate the given number.

Mathematics
1 answer:
bagirrra123 [75]3 years ago
4 0
Would use the algorithm for solving square root.

For square root,  √n

x₁ = 0.5(x₀ + n/x₀)        

(This formula is known and for square root, and can be derived using Newton-Raphson's approximation equation)

Where x₀ is the initial guess.  x₁ becomes the new guess.

For √100.6  let our initial guess be 10,  x₀ = 10,  n = 100.6

Our approximation shall be to 3 decimal places. Once we get the same answer twice we stop the algorithm.

x₀ = 10,  x₁ = 0.5(x₀ + n/x₀),    x₁ = 0.5(10 + 100.6/10)  = 10.030, x₁ = 10.030

x₂ = 0.5(x₁ + n/x₁),    x = 0.5(10.030 + 100.6/10.030)  ≈10.015,  x₂ ≈ 10.030 (to 3 decimal places)

Since x₂≈ x₁, the algorithm stops.

So the √100.6 is ≈ 10.030 to 3 decimal places.

I hope this helps.
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Mal spends 7 hours in school each day. Her lunch period is 1 hour long, and she spends a
Alex787 [66]

Answer:

z = 5*(1/2)

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---

time switching classes:

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---

y - 6x - z - w = 0

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x = (76 - 5 - 7)/(10*6)

x = (64)/(10*6)

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check:

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y = 7.6

z = 1/2

z = 0.5

w = 7/10

w = 0.7

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (7.6 - 0.5 - 0.7)/6

x = 1.0666666666

---

answer:

z = 5*(1/2)

z = 5/10

---

time switching classes:

w = 7/10

---

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (76/10 - 5/10 - 7/10)/6

x = (76 - 5 - 7)/(10*6)

x = (64)/(10*6)

x = (2*2*2*2*2*2)/(2*5*2*3)

x = (2*2*2*2)/(5*3)

x = 16/15

x = 1.0666666666

---

check:

y = 7 + 3/5

y = 7.6

z = 1/2

z = 0.5

w = 7/10

w = 0.7

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (7.6 - 0.5 - 0.7)/6

x = 1.0666666666

---

answer:

each class is 1.07 hours

Step-by-step explanation:

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