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stealth61 [152]
3 years ago
11

2y/(z-y)x=4y solve for x

Mathematics
1 answer:
alexdok [17]3 years ago
3 0

Answer:

  x = 2(z -y)

Step-by-step explanation:

Multiply by the inverse of the coefficient of x.

  x = 4y(z -y)/2y

  x = 2(z -y) . . . . . . . . y ≠ 0

_____

The equation is meaningless if y = 0.

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The simplest polynomial function with zeros 7, square root 13 and 27i would be of what degree
Harman [31]

Answer:

The simplest polynomial function is y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}.

Step-by-step explanation:

A n-th order polynomial in factorized form is defined by:

y = (x-r_{1})\cdot (x-r_{2})\cdot ...\cdot (x-r_{n-1})\cdot (x-r_{n}) (1)

Where:

x - Independent variable.

y - Dependent variable.

r_{1}, r_{2},..., r_{n-1}, r_{n} - Roots of the polynomial.

We know that three roots, two real and a complex root. By Quadratic Formula,  a second order polynomial has two conjugated complex number of the form a + i\,b and a - i\,b. Hence, there is an additional zero: -i\,27.

By applying (1), we have the following polynomial when all roots are used:

y = (x-7)\cdot (x-\sqrt{13})\cdot (x-i\,27)\cdot (x+i\,27)

By factorization and algebraic handling we have the simplest polynomial function:

y = [x^{2}-(7+\sqrt{13})\cdot x +7\sqrt{13}]\cdot (x^{2}+27)

y = x^{2}\cdot (x^{2}+27)-(7+\sqrt{13})\cdot [(x^{2}+27)]\cdot x+7\sqrt{13}\cdot (x^{2}+27)

y = x^{4}+27\cdot x^{2}-(7+\sqrt{13})\cdot x^{3}-(7+\sqrt{13})\cdot 27\cdot x +7\sqrt{13}\cdot x^{2}+189\sqrt{13}

y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}

The simplest polynomial function is y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}.

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3 years ago
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