MaxWong

Component form of a two-dimensional vector is (a, b), where a and b are real numbers.

First, we want to know what does (-2, 3) mean. It means the direction "two units to the left, and three units upwards".

The result of multiplying this vector by 6 means that going "two units to the left and three units upwards" 6 times.

In total, you would have gone 12 units to the left and 18 units upwards, which is (-12, 18). 

You should be able to choose the same option for two different blanks.

First, the angle of rotation -pi/2 means a clockwise rotation of pi/2 radians from the positive x-axis. That would give the negative y-axis. So if r is positive, then the point lies on negative y-axis and if r is negative, then the point lies on positive y-axis. 

In this case, r = -5, which means the point lies on the positive y-axis. 

Note that not all points with negative r lies on negative x- or y-axes.

(1, 1, 1) and (-2, -2, -2) are linearly dependent vectors, which means all points on (x, y, z) = (1, 2, 3) + s(1, 1, 1) + t(-2, -2, -2) all satisfies (x - 1, y - 2, z - 3) = (s - 2t) (1, 1, 1), which means it is a straight line passing through the point (1, 2, 3).

(x, y, z) = (1, 2, 3) + s(1, 1, 1) + t(-2, -2, -2)

(x, y, z) = (1, 2, 3) + s(1, 1, 1) - 2t(1, 1, 1)

(x, y, z) = (1, 2, 3) + (s - 2t) (1, 1, 1)

It is an equation of a line.

The range is (-1, 0].

The graph: https://www.desmos.com/calculator/e2oq2ajmsi

More generally, we consider the sum \(S = \displaystyle\sum_{n = 0}^\infty F_n x^n\), \(|x| < \dfrac{\sqrt 5 - 1}2\).

Here, the case is x = 1/4, but we will consider that later.

By Binet's formula, \(F_n = \dfrac{(1 + \sqrt 5)^n - (1 - \sqrt 5)^n}{2^n \sqrt 5}\).

Substituting this into the sum and denote \(\dfrac{1 + \sqrt 5}2 \) as \(\varphi\),

\(S = \displaystyle\sum_{n = 0}^\infty F_n x^n = \dfrac1{\sqrt 5} \sum_{n = 0}^\infty \left((\varphi x)^n - \left(\dfrac{-x}{\varphi}\right)^n\right)\)

This is the sum of G.S., and can be calculated by the formula \(\displaystyle\sum_{n = 0}^\infty r^n = \dfrac{1}{1 - r}\).

Therefore \(S = \dfrac1{\sqrt{5}} \left(\dfrac1{1 - x\varphi} - \dfrac1{1 + \dfrac{x}{\varphi}}\right)\).

Now, substituting x = 1/4,

\(S = \dfrac1{\sqrt 5} \left(\dfrac1{1 - \dfrac{\varphi}4} - \dfrac1{1 + \dfrac1{4\varphi}}\right)\)

Upon simplification, we have 

\(S = \dfrac1{\sqrt 5} \left(\dfrac{4\sqrt 5}{11}\right) = \boxed{\dfrac 4{11}}\)

The computation I did with Wolfram: https://www.wolframalpha.com/input/?i=1%2F%281+-+%28golden+ratio%29%2F4%29+-+1%2F%281+%2B+1%2F%284%28golden+ratio%29%29%29

Let \(t = \sqrt x\). 

\(t^2 = x\)

So the objective function is just \(t - t^2\).

Completing the square yields \(t - t^2 = \dfrac14 - \left(t - \dfrac12\right)^2\)

So, when \(t = \dfrac12\), the maximum value of the objective function is attained, and the value is \(\boxed{\dfrac14}\)

Yes, you have converted them correctly. But do you mean \(\tan\left(\theta + \varphi\right)\) instead? Because you asked how to add theta and phi together.

Here is how you add theta and phi. You first convert the arccos in theta to arctan with trigonometry. Then you can use the formula \(\arctan A + \arctan B = \arctan \dfrac{A + B}{1 - AB}\)

I am here to provide a fun alternative solution.

x + y = xy = 3 means that x and y are the roots of the quadratic equation \(t^2 - 3t + 3 = 0\).

(You can get this result by using Vieta's formula.)

Now, substituting x and y into the equation, \(\begin{cases}x^2 - 3x + 3 = 0\\y^2 - 3y + 3=0\end{cases}\), which means \(\begin{cases}x^2 = 3(x - 1)\\y^2 = 3(y - 1)\end{cases}\).

Using the result above:

\(\quad x^3 + y^3 \\= x(\color{red}x^2\color{black}) + y(\color{blue}y^2\color{black}) \\= \color{red}3\color{black}x\color{red}(x - 1)\color{black} + \color{blue}3\color{black}y\color{blue}(y - 1)\color{black} \\= 3x^2 + 3y^2 - 3(x + y)\)

Substituting x^2 = 3(x - 1) and y^2 = 3(y - 1) again and using the fact that x + y = 3,

\(\quad x^3 + y^3\\ = 3\color{red}x^2\color{black} + 3\color{blue}y^2\color{black} - 3(x + y)\\ = 3(\color{red}3(x - 1)\color{black}) +3(\color{blue}3(y - 1)\color{black}) - 3(x+y)\\ = 9x - 9 + 9y - 9 - 3(x + y) \\= 9(x + y) - 18 - 3(x + y)\\ =6(\color{purple}x + y\color{black}) - 18\\ =6(\color{purple}3\color{black}) - 18\\ =0\)

Correct. That's because \(0.64 = \dfrac{64}{100} = \dfrac{8^2}{10^2} = \left(\dfrac{8}{10}\right)^2 = (0.8)^2\).