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ddd [48]
3 years ago
10

Wire is commonly made by putting hot metal through a hole in a plate. What length of wire, of diameter 1mm, can be made from a 1

cm cube of metal?
Mathematics
1 answer:
shtirl [24]3 years ago
7 0
Basing on the question the volume of metal should be equal to the volume of the wire.
We already have the volume of the metal which is 1cm cube.
To get the volume of the wire, the equation is V=3.14 x r^2 x h
since the volume of the wire is same as the volume of the metal, we simply substitute.

1 cm^3 = 3.14 x r^2 x h

h is the length of the wire.
the diameter of the wire is 1 mm or 0.1 cm, to get the radius we divide it by 2, 0.1 cm / 2 is 0.05 cm, we substitute

1 cm^3= 3.14 x (0.05 cm)^2 x h
1 cm^3= 3.14 x 0.0025 cm^2 x h
h = 0.00785 cm^2 / 1 cm^3
h=0.00785 cm or 0.0785 mm.
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Using the quadratic formula, what is the value of x for x^2-4+5=0​
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Step-by-step explanation:

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Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
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The nth taylor polynomial for the given function is

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Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

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f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

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f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

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P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

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