Featured 1 month ago

Please see the explanation.

The slope, m, of the line is:

Use the point-slope form of the equation of a line:

Here is a graph of that line:

Multiply both sides by 3:

Distribute the -2:

Add 2x to both sides:

Substitute

Factor out r:

Divide both sides by

Here is a graph of that equation:

Featured 1 month ago

Let's first notice that we're looking at

There are

We need to go around once more time:

Now let's go around a third time, but not all the way. We'll go through Q4 (to do that takes

Do you see the point for

Now that we know where we are, we can get down to finding the cosine!

The reference angle between

One last thing to address and we're done - because the adjacent is along the negative

Featured 1 month ago

See the **Proof** in the **Explanation.**

Given that,

Similarly,

Hence, **Squaring and Adding**

Featured 1 month ago

Use Newton method to find a zero of

The iterations approach

Featured 3 weeks ago

So first up we need to understand the question completely. A helicopter starts at

So first we need to find the velocity vector which would be

To find the vector between two points

So from our points, we have:

This vector is the velocity vector of the helicopter in terms of km per 10 minutes (as it travels from

This vector would look like:

(the red dot is A and the Blue dot is B)

So we need to find the angle that

To do this find the angle between

To find the angle between the two vectors

In this equation, the numerator is the modulus of the dot product and the denominator is the length of each vector.

So when we are finding the angle between

Featured 3 weeks ago

Using formula

Featured 3 weeks ago

Q-1

Let

Again

So

**Otherwise without using exact values**

Hence

Now

Q-2

Now

So

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

Featured 2 weeks ago

As per problem the movement of the London Eye is periodic one and its height at

*meter* and t in *sec*.

Here *cycle* is 3o min. This means time period

Hence

So the equation[1] takes the following form

Now at

Applying this condition on [2] we get

Since at

then we can say that at

So we can write

Now adding [3] and [4] we have

Subtracting [4] from [3] we get

So finally the given equation takes the following form

The variation of height with time can be represented by following graph.

Featured 18 hours ago

Use:

#sin 30^@ = 1/2#

#cos 30^@ = sqrt(3)/2#

#sin 45^@ = cos 45^@ = sqrt(2)/2#

#sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha#

#cos(alpha-beta) = cos alpha cos beta + sin alpha sin beta#

#tan theta = sin theta / cos theta#

Then:

#sin 15^@ = sin (45^@ - 30^@)#

#color(white)(sin 15^@) = sin 45^@ cos 30^@ - sin 30^@ cos 45^@#

#color(white)(sin 15^@) = sqrt(2)/2 sqrt(3)/2 - 1/2 sqrt(2)/2#

#color(white)(sin 15^@) = sqrt(2)/4(sqrt(3)-1)#

#cos 15^@ = cos (45^@ - 30^@)#

#color(white)(cos 15^@) = cos 45^@ cos 30^@ + sin 45^@ sin 30^@#

#color(white)(cos 15^@) = sqrt(2)/2 sqrt(3)/2 + sqrt(2)/2 1/2#

#color(white)(cos 15^@) = sqrt(2)/4(sqrt(3)+1)#

#tan 15^@ = sin 15^@ / cos 15^@#

#color(white)(tan 15^@) = (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)-1)) / (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)+1))#

#color(white)(tan 15^@) = (sqrt(3)-1) / (sqrt(3)+1)#

#color(white)(tan 15^@) = ((sqrt(3)-1)^2) / ((sqrt(3)+1)(sqrt(3)-1))#

#color(white)(tan 15^@) = (3-2sqrt(3)+1) / (3-1)#

#color(white)(tan 15^@) = 2 - sqrt(3)#

**Footnote**

We can see the original values of