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Volgvan
3 years ago
6

A wire 390 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures ha

ve the same area, what are the lengths of the two pieces of wire (to the nearest tenth of an inch)?

Mathematics
2 answers:
kupik [55]3 years ago
7 0

Step-by-step explanation:

Below is an attachment containing the solution.

jeka57 [31]3 years ago
3 0

Answer:

Step-by-step explanation:

Let x is the length of the square side.

We know: the two figures have the same area

<=> x^{2} = πr^{2} (r being radius)

<=> x = \sqrt{II} r

perimeter square = 4x = 4*r*\sqrt{II}

perimeter circle = 2*r*π

<=> 390 = 4*r*\sqrt{II}  + 2*r*π

<=> r = 29.16

=> perimeter square =  4*29.16*π

=> perimeter circle = 58.32π

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Given that W is directly proportional to the square of V, and W = 320 when V = 8, find W when V = 9.
lorasvet [3.4K]

The value of W when V = 9 is 405

<h3>How to solve for W?</h3>

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2 years ago
Determine for which values of m the function variant Φ(x) = x^m is a solution to the given equation. a. 3x^2 (d^2y/dx^2) + 11x(d
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b) m =  1 + √6 or m = 1 -√6

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for

Φ(x) = x^m

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dΦ/dx (x) = m*x^(m-1)

d²Φ/dx² (x) = m*(m-1)*x^(m-2)

then

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3x^2*m*(m-1)*x^(m-2) + 11*x* m*x^(m-1) - 3*x^m = 0

3*m*(m-1)*x^m + 11*m*x^m- 3*x^m = 0

dividing by x^m

3*m*(m-1) + 11*m - 3 =0

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for b)

x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

x^2*m*(m-1)*x^(m-2) -  x* m*x^(m-1) - 5*x^m = 0

m*(m-1)*x^m -m *x^m- 5*x^m = 0

dividing by x^m

m*(m-1) -m - 5 =0

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then Φ(x) = x^m is a solution for the equation b , when m =  1 + √6 or m = 1 - √6

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