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2 years ago

11

I need to know how to do these

1 answer:

kenny6666 [7]2 years ago

5 0

Answer

Step-by-step explanation:

ur gonna have to divide acctually to get the decimal then multipy the decimal and payment

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The equation of the tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 at the point (x0, y0, z0) can be written as xx0 a2

svetlana [45]

Answer:

The equation of tangent plane to the hyperboloid

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

Step-by-step explanation:

Given

The equation of ellipsoid

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

The equation of 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 equation of hyperboloid

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

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

$F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}$

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

The equation of tangent plane 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 equation of tangent plane to the hyperboloid

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

The equation of tangent plane

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

Hence, the required equation of tangent plane to the hyperboloid

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

7 0

3 years ago

3x+2=11 plz hep im slow in math

stealth61 [152]

Answer:

3

Step-by-step explanation:

3×3=9+2=11. so 3 is the answer.

3 0

3 years ago

Please help me Algebra Graph!

Luba_88 [7]

There is no attachment to your phone question. I look forward to answer your question.

3 0

2 years ago

Is this right or wrong ?

Otrada [13]

Answer:

$(3x - 2)(4x + 6) = 12 {x}^{2} + 18x - 8x + - 12 \\ = 12 {x}^{2} + 10x - 12$

6 0

3 years ago

What is the difference in Fahrenheit degrees between a temperature of 0C and a temperature of 15C?

slamgirl [31]

Answer:

$27^{\circ}F$

Step-by-step explanation:

Use formula

$F=1.8C+32$

to convert Celsius to Fahrenheit.

When $C_1=0^{\circ}C,$ then

$F_1=1.8\cdot 0+32\\ \\F_1=32^{\circ}F$

When $C_2=15^{\circ}C,$ then

$F_2=1.8\cdot 15+32\\ \\F_2=27+32=59^{\circ}F$

Difference:

$F_2-F_1=59^{\circ}F-32^{\circ}F=27^{\circ}F$

8 0

3 years ago