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Alborosie
3 years ago
9

If f(1) = 160 and f(n + 1) = –2f(n), what is f(4)?

Mathematics
2 answers:
lions [1.4K]3 years ago
5 0

Answer:

-1280

Step-by-step explanation:

There are 2 ways you could do this. You could just do the question until you come to the end of f(4). That is likely the simplest way to do it.

f(1) = 160

f(2) = - 2 * f(1)

f(2) = -2*160

f(2) = -320

f(3) = -2 * f(2)

f(3) = -2 * - 320

f(3) = 640

f(4) = - 2 * f(3)

f(4) = - 2 * 640

f(4) = - 1280

I don't know that you could do this explicitly with any real confidence.

Gre4nikov [31]3 years ago
3 0

f(1)=160\\f(n+1)=-2f(n)\\\\f(2)=-2\cdot 160=-320\\f(3)=-2\cdot(-320)=640\\f(4)=-2\cdot 640=-1280

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Answer:

No

Step-by-step explanation:

This is not a binomial situation, as there are more than two outcomes (the prefix bi means two). There is also not an equal amount of chance of pulling any color marble.

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3 years ago
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
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Answer:

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Step-by-step explanation:

if you want an explanation, please, dont be afraid to comment ;)

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