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3 years ago

A ____ helps plan for retirement.

Mathematics

1 answer:

VMariaS [17]3 years ago

6 0

Answer:

the answer would be A :) just took the test

Step-by-step explanation:

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Using a deck of 52 playing cards, what are the odds in favor of drawing a Jack or 10?

andrey2020 [161]

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3 years ago

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3 years ago

Read 2 more answers

Question is the attached file.

Gwar [14]

Answer:

The critical points are 12 and 0.

Step-by-step explanation:

We have that the critical numbers are those values that result from equating the derivative of a function to zero. Also called roots or zeros of the derived function.

IF f is defined in x, it will be said that a is a critical number of f if f '(x) = 0 or if f is not defined in x.

Now the function is:

f (x) = x ^ 2 / (x -6)

we have that the derivative of the quotient is:

(f / g) '= (f' * g - g '* f) / g ^ 2

we replace and we have:

f (x) = [2 * x * (x-6) - 1 * x ^ 2] / (x -6) ^ 2

simplifying we have:

f (x) = [x ^ 2 - 12 * x] / (x -6) ^ 2

this must be equal to 0, like this:

[x ^ 2 - 12 * x] / (x -6) ^ 2 = 0

we solve:

x ^ 2 - 12 * x = 0

x * (x - 12) = 0

Thus:

x = 0

x - 12 = 0 => x = 12

The critical points are 12 and 0.

6 0

3 years ago

Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect

Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

$\bf (cos(\pi t), sin(\pi t), t)$

But

$\bf cos^2(\pi t)+sin^2(\pi t)=1$

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface

$\bf z=x^2+y^2$

at the point (-1, 0, 1) is given by

$\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)$

where

$\bf f(x,y)=x^2+y^2$

since

$\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y$

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>is given by

$\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)$

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.

But we now

$\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta$