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Vaselesa [24]
3 years ago
15

What is the formula for the area of a triangle and a square again?

Mathematics
2 answers:
Tatiana [17]3 years ago
5 0

1/2 X base X height= triangle

length X width X height= square

horsena [70]3 years ago
3 0

\bf \begin{array}{|c|ll} \cline{1-1} \textit{area of a triangle}\\ \cline{1-1} \\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height \end{cases} \\\\ \cline{1-1} \end{array}~\hspace{7em}  \begin{array}{|c|ll} \cline{1-1} \textit{area of a square}\\ \cline{1-1} \\ A=s^2\qquad  \begin{array}{llll} s=length~of\\ \qquad a~side \end{array} \\\\ \cline{1-1} \end{array}

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Answer:

No solution

Step-by-step explanation:

Given equation is,

\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}-\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}=0

\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}=\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}

\frac{(x+1)}{\sqrt{x}(1-x)}+\frac{(\sqrt{x}-1)}{\sqrt{x}(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}

\frac{(\sqrt{x}+1)(x+1)+(\sqrt{x}-1)(1-x)}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}

\frac{x\sqrt{x}+x+\sqrt{x}+1+\sqrt{x}-1-x\sqrt{x}+x}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}

\frac{2x+2\sqrt{x}}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}

\frac{2(\sqrt{x}+1)}{(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}

\frac{2}{1-x}=(\frac{4+x}{1-x})^\frac{1}{2}  if x ≠ ±1

(\frac{2}{1-x})^2=\frac{4+x}{1-x}  [Squaring on both the sides of the equation]

\frac{4}{(1-x)}=(4+x)

4 = (1 - x)(4 + x)

4 = 4 - 4x + x - x²

0 = -3x - x²

x² + 3x = 0

x(x + 3) = 0

x = 0, -3

But both the solutions x = 0 and x = -3 are extraneous solutions, given equation has no solution.

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