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lakkis
2 months ago
9

3x+8+2ax≥3ax−4a what is x?

Mathematics
2 answers:
zzz [12.3K]2 months ago
7 0
X≥
- \frac {4 (2 + a)}{3 - a}
to resolve the inequality, identify the roots and establish testing intervals.
zzz [12.3K]2 months ago
6 0

Response:

\text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}

Step-by-step breakdown:

Given: the inequality 3x+8+2ax≥3ax−4a

Goal: to determine the value of x

Solution:

within the provided inequality

3\text{x}+8+2\text{a}\text{x}\geq 3\text{a}\text{x}-4\text{a}

3\text{x}+2\text{a}\text{x}-3\text{a}\text{x}\geq -4\text{a}-8

\text{x}(3+2\text{a}-3\text{a})\geq -8-4\text{a}

\text{x}(3-\text{a})\geq -8-4\text{a}

\text{x}\geq \frac{-8-4\text{a}}{(3-\text{a})}

\text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}

thus the value of \text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}

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\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1.

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The ellipsoid's equation is

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The equation for the tangent plane at the point \left(x_0,y_0,z_0\right)

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1  (Given)

The hyperboloid's equation is

\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1

F(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}

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(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)

The tangent plane equation at point \left(x_0,y_0,z_0\right)

\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0

The tangent plane equation for the hyperboloid is

\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0

The tangent plane equation

2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2

Hence, the required tangent plane equation for the hyperboloid is

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0

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