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Featured 5 months ago

Here's a rough method...

Here's one naive and slightly long-winded approach, which may or may not be the sort of thing the posers of the question had in mind...

Given:

We want to match this to:

#f(x) = A sin(omega x - phi) + beta#

where

The natural period of

So

Note that

#(6.21, 3.91)# ,#(7.02, 4.36)# ,#(7.84, 5.31)# ,

#(8.19, 6.21)# ,#(8.06, 7.02)# ,#(7.41, 7.84)# ,

#(6.30, 8.19)# ,#(5.21, 8.06)# ,#(4.28, 7.41)# ,

#(3.91, 6.30)# ,#(4.36, 5.21)# ,#(5.31, 4.28)#

The centre of the circle is

#m=1/12(3.91+4.36+5.31+6.21+7.02+7.84+8.19+8.06+7.41+6.30+5.21+4.28) = 6.175#

If all of the

The distance between

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Putting

#2.27# ,#2.00# ,#1.88# ,#2.02# ,#2.07# ,#2.07# ,#2.02# ,#2.12# ,#2.26# ,#2.27# ,#2.06# ,#2.08#

with mean:

So far, we have determined

Notice that

So we might as well approximate the minimum of the sinusoid at

So our formula is:

#f(x) = A sin(omega x - phi) + beta = 2.09 sin(pi/6 x - (2pi)/3) + 6.175#

which results in monthly values:

#4.09# ,#4.37# ,#5.13# ,#6.17# ,#7.22# ,#7.98# ,

#8.27# ,#7.98# ,#7.22# ,#6.17# ,#5.13# ,#4.37#

graph{(y-(2.09 sin(pi x / 6-2pi/3) + 6.175))((x-1)^2+(y-3.91)^2-0.03)((x-2)^2+(y-4.36)^2-0.03)((x-3)^2+(y-5.31)^2-0.03)((x-4)^2+(y-6.21)^2-0.03)((x-5)^2+(y-7.02)^2-0.03)((x-6)^2+(y-7.84)^2-0.03)((x-7)^2+(y-8.19)^2-0.03)((x-8)^2+(y-8.06)^2-0.03)((x-9)^2+(y-7.41)^2-0.03)((x-10)^2+(y-6.30)^2-0.03)((x-11)^2+(y-5.21)^2-0.03)((x-12)^2+(y-4.28)^2-0.03) = 0 [-0.85, 19.15, 1.56, 11.56]}

Featured 5 months ago

Given:

focus:

directrix:

Rewrite the equation for the directrix as:

This is done to that it fits the formula for the distance from a point to a line:

where

The distance from the focus,

A parabola is the locus of points that are equidistant from its focus and its directrix, therefore, we can derive the equation of the parabola by setting the right side of equation [1.2] equal to the right side of equation [2.1]:

Multiply both sides by

Square both sides:

Expand the square:

Combine like terms and write the equation in the General Cartesian form of a conic section ,

Here is a graph of the parabola, the focus and the directrix:

Featured 5 months ago

We can prove this by contradiction;

if

so hence let,

then

so hence

we know

so hence

so hence

so

But we can see that

Featured 4 months ago

For each row of the first matrix and column of the second, sum the products of the three corresponding terms...

#((color(red)(1), color(red)(-3), color(red)(2)), (2, 1, -3), (4, -3, -1))((color(blue)(2), 1, -1, -2),(color(blue)(3), -2, -1, -1),(color(blue)(2), -5, -1, 0)) = ((color(purple)(-3), ?, ?, ?),(?, ?, ?, ?), (?, ?, ?, ?))#

#(color(red)(1))(color(blue)(2))+(color(red)(-3))(color(blue)(3))+(color(red)(2))(color(blue)(2)) = 2-9+4 = color(purple)(-3)#

#((color(red)(1), color(red)(-3), color(red)(2)), (2, 1, -3), (4, -3, -1))((2, color(blue)(1), -1, -2),(3, color(blue)(-2), -1, -1),(2, color(blue)(-5), -1, 0)) = ((-3, color(purple)(-3), ?, ?),(?, ?, ?, ?), (?, ?, ?, ?))#

#(color(red)(1))(color(blue)(1))+(color(red)(-3))(color(blue)(-2))+(color(red)(2))(color(blue)(-5)) = 1+6-10 = color(purple)(-3)#

#dot dot dot#

#((1, -3, 2), (color(red)(2), color(red)(1), color(red)(-3)), (4, -3, -1))((color(blue)(2), 1, -1, -2),(color(blue)(3), -2, -1, -1),(color(blue)(2), -5, -1, 0)) = ((-3, -3, 0, 1),(color(purple)(1), ?, ?, ?), (?, ?, ?, ?))#

#(color(red)(2))(color(blue)(2))+(color(red)(1))(color(blue)(3))+(color(red)(-3))(color(blue)(2)) = 4+3-6 = color(purple)(1)#

#dot dot dot#

#((1, -3, 2), (2, 1, -3), (color(red)(4), color(red)(-3), color(red)(-1)))((2, 1, -1, color(blue)(-2)),(3, -2, -1, color(blue)(-1)),(2, -5, -1, color(blue)(0))) = ((-3, -3, 0, 1),(1, 15, 0, -5), (-3, 15, 0, color(purple)(-5)))#

#(color(red)(4))(color(blue)(-2))+(color(red)(-3))(color(blue)(-1))+(color(red)(-1))(color(blue)(0)) = -8+3+0 = color(purple)(-5)#

Featured 3 months ago

First find

The easiest way to find the inverse of

This is just:

Next switch the elements on the leading diagonal of

So you should have:

Divide each element by the determinant

Now:

Using

**Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.**

i.e.

**( In general )**

Featured 3 months ago

Let

What is

Subtracting latter from former, we get

or

=

=

=

=

and

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