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Featured 6 months ago

Alternate solution for part (c)

For Solution posted by @dk_ch

The modelling equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for

As such the stock actually lost value from May to August.

--.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

For Solution posted by @Cesareo R

#f(t)=2.25t+18.49+7.29sin((pit)/6)# , values rounded to two decimal places

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for

As such the stock actually lost value from May to August.

Featured 5 months ago

Use the cosine angle addition formula:

#cos(x+y)=cosxcosy-sinxsiny#

We need to determine

**Determining**

#tan^2x+1=sec^2x" "" "" "# use#tanx=5/3#

#25/9+1=sec^2x#

#secx=sqrt(34/9)=sqrt34/3#

Then:

#color(blue)(cosx=1/secx=3/sqrt34#

Now using:

#sin^2x+cos^2x=1" "" "" "# where#cosx=3/sqrt34#

#sin^2x+9/34=1#

#color(blue)(sinx=sqrt(25/34)=5/sqrt34#

**Determining**

#sin^2y+cos^2y=1" "" "" "# use#siny=1/3#

#1/9+cos^2y=1#

#color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3#

**Returning to**

#cos(x+y)=cosxcosy-sinxsiny#

#color(white)(cos(x+y))=3/sqrt34((2sqrt2)/3)-5/sqrt34(1/3)#

#color(white)(cos(x+y))=(6sqrt2-5)/(3sqrt34)#

Note these are all assuming that

Featured 3 months ago

Period of

#f(x) = a \sin(bx+c) + d# ,

where

Ignoring the shifts along axes i.e. assuming

#f(x) = y=1+4 \sin(2x)#

#\Rightarrow y -1 = 4 \sin(2x)#

#\Rightarrow g(x) = 4 \sin(2x)#

We know from fundamental trigonometric ideas that the period of sine function is

#\sin(x) = \sin(2n \pi +x)# ,

Let,

#\sin(u) = \sin(2n \pi + u)#

Then for

#\sin(2x) = \sin(2n \pi +2x)#

Multiplying both sides by

#4 \sin(2x) = 4\sin 2(n \pi + x)#

#\equiv g(x) = g(n\pi+x)#

Hence period is

In general the amplitude of sine function is given as;

If

graph{y=\sin(x)[-3,3,-1.5,1.5]}

Our equation is a sort of

Which becomes;

#y= a[\sin(bx)]#

And that expands the graph of ordinary sine function i.e.

graph{y=4\sin(2x)}

Therefore our amplitude is

The graph of our real function

graph{y=1+4 \sin(2x)}

But amplitude remains same.

Featured 3 months ago

(a)

[Taking

Where

(b)

(i)

The value of

So

(ii)Here

Featured 1 month ago

Please see below.

It isassumed that in the equation **sine** ratio of an angle always lies between

and

Also cosider two equations

graph{(y-sinx)(y-0.75)=0 [-1.25, 8.75, -2.48, 2.52]}

graph{(y-sinx)(y+0.5)=0 [-1.25, 8.75, -2.48, 2.52]}

Observe that each has two solutions. In first case when

Similarly, solution for

Hence, if **sine** is

If **sine** is

General solutions are

Featured 2 weeks ago

A few thoughts...

It was known and studied by Euclid (approx 3rd or 4th century BCE), basically for many geometric properties...

It has many interesting properties, of which here are a few...

The Fibonacci sequence can be defined recursively as:

#F_0 = 0#

#F_1 = 1#

#F_(n+2) = F_n + F_(n+1)#

It starts:

#0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,...#

The ratio between successive terms tends to

#lim_(n->oo) F_(n+1)/F_n = phi#

In fact the general term of the Fibonacci sequence is given by the formula:

#F_n = (phi^n - (-phi)^(-n))/sqrt(5)#

A rectangle with sides in ratio

This is related to both the limiting ratio of the Fibonacci sequence and the fact that:

#phi = [1;bar(1)] = 1+1/(1+1/(1+1/(1+1/(1+1/(1+...)))))#

which is the most slowly converging standard continued fraction.

If you place three golden rectangles symmetrically perpendicular to one another in three dimensional space, then the twelve corners form the vertices of a regular icosahedron. Hence we can calculate the surface area and volume of a regular icosahedron of given radius. See https://socratic.org/s/aFZyTQfn

An isosceles triangle with sides in ratio

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