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bonufazy [111]
3 years ago
13

- 4x4 – 13x3 + 8x2 – 21x + 18 is divided by 2 – 3?

Physics
2 answers:
Alecsey [184]3 years ago
7 0

Answer:

x = 3/4 or x = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3 or x = 1/3 (13 (-1/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) - 4/3 or x = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3

Explanation:

Solve for x over the real numbers:

4 x^4 + 13 x^3 - 8 x^2 + 21 x - 18 = 0

The left hand side factors into a product with two terms:

(4 x - 3) (x^3 + 4 x^2 + x + 6) = 0

Split into two equations:

4 x - 3 = 0 or x^3 + 4 x^2 + x + 6 = 0

Add 3 to both sides:

4 x = 3 or x^3 + 4 x^2 + x + 6 = 0

Divide both sides by 4:

x = 3/4 or x^3 + 4 x^2 + x + 6 = 0

Eliminate the quadratic term by substituting y = x + 4/3:

x = 3/4 or 14/3 + 4 (y - 4/3)^2 + (y - 4/3)^3 + y = 0

Expand out terms of the left hand side:

x = 3/4 or y^3 - (13 y)/3 + 254/27 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

x = 3/4 or 254/27 - 13/3 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

x = 3/4 or z^6 + z^4 (3 λ - 13/3) + (254 z^3)/27 + z^2 (3 λ^2 - (13 λ)/3) + λ^3 = 0

Substitute λ = 13/9 and then u = z^3, yielding a quadratic equation in the variable u:

x = 3/4 or u^2 + (254 u)/27 + 2197/729 = 0

Find the positive solution to the quadratic equation:

x = 3/4 or u = 1/27 (18 sqrt(43) - 127)

Substitute back for u = z^3:

x = 3/4 or z^3 = 1/27 (18 sqrt(43) - 127)

Taking cube roots gives 1/3 (18 sqrt(43) - 127)^(1/3) times the third roots of unity:

x = 3/4 or z = 1/3 (18 sqrt(43) - 127)^(1/3) or z = -1/3 (-1)^(1/3) (18 sqrt(43) - 127)^(1/3) or z = 1/3 (-1)^(2/3) (18 sqrt(43) - 127)^(1/3)

Substitute each value of z into y = z + 13/(9 z):

x = 3/4 or y = 1/3 (18 sqrt(43) - 127)^(1/3) - (13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) or y = 13/3 ((-1)/(127 - 18 sqrt(43)))^(1/3) - 1/3 (-1)^(1/3) (18 sqrt(43) - 127)^(1/3) or y = 1/3 (-1)^(2/3) (18 sqrt(43) - 127)^(1/3) - 13/(3 (127 - 18 sqrt(43))^(1/3))

Bring each solution to a common denominator and simplify:

x = 3/4 or y = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) or y = 1/3 (13 ((-1)/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) or y = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3))

Substitute back for x = y - 4/3:

Answer: x = 3/4 or x = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3 or x = 1/3 (13 (-1/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) - 4/3 or x = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3

torisob [31]3 years ago
7 0

Answer:

-58 i think

Explanation:

-16 - 39 + 16 - 21x = 18 divid 2 -3

-16 - 55 - 39x divided by 2 - 3

-110 divided by 2 - 3

-55 - 3 = -58

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Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

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Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

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Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

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Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

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