Featured 1 month ago

Given that

Let us consider

so

From (2) and (3) we get

Hence

**(a)**

**(b)**

Now the given equation becomes

Differentiating this equation w,r to t we get the rate of change of the price of stock as

In this equation the value of

So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

**(c)**

Taking

when t =4

Again

when t =8

So during the period **May -September** growth of price gets diminished i.e the price of the stock is actually **losing in this period.**

Featured 1 month ago

The equation

*Graph has been plotted taking 12 O clock night as zero hour i,e, t = 0*

The condition to dive off safely is that the water must be at least

So inserting

when n =0

when n =1

and

Hence from above calculation as well as from graphical presentation of variation of depth of water with time it is obvious that **during daytime** the water level remains at least 3m deep for the period

and this period is safe to dive off .

Featured 3 weeks 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 4 weeks ago

You have to start by finding a point that lies on this line. I think the simplest would be

We now draw an imaginary triangle, as shown in the following diagram.

We can now clearly see that our side opposite

We must finally find the hypotenuse prior to determining the ratios. By pythagorean theorem:

However, the hypotenuse can never have a negative length, so we can only accept the positive solution. We can now find our ratios.

Hopefully this helps!

Featured 3 weeks ago

See below:

With a triangle with

We could run through the pythagorean theorem for the third side, but we can also remember that these two measurements are part of a 30-60-90 triangle, and so the third side is

This gives us the 6 trig ratios:

The angle of the triangle we're looking at is

Featured 1 week ago

The angle

When we talk about the quadrant an angle is in, we're really talking about the quadrant the *terminal arm* of the angle is in, when the angle is drawn with its initial arm along the

To draw an angle, imagine a clock centered at the origin, with the minute hand pointing to the right (at the "3"). Then, rotate the hand counterclockwise (towards the "12") by the amount of the angle. Since the angle

Now,

Which quadrant is this? The quadrants are named in increasing order, starting with quadrant 1 (

So the angle

Featured 1 week ago

Note that

In this case

"Which angle has a

The only way to determine this is with a calculator or tables.

Using a graph is possible, but not accurate enough.

Depending on which calculator you have, the following are possible key presses:

D.A.L. calculators:

shift

The answer will be

Or

The answer will be

Note that this is only the acute angle, the 1st quadrant angle.

In the 4th quadrant,

Featured 2 weeks ago

See below:

To help follow what's going on I'll put in red text things that have changed from the line before.

I'll first use

We can now use

Featured 1 week ago

If

then

and

If

then

and

Plugging in the values for

If we call this value

then

and (since