Distances in 2- and 3-dimensions (and even higher dimensions) can be found using the Pythagorean theorem. The straight-line distance can be considered to be the hypotenuse of a right triangle whose sides are the horizontal and vertical differences between the coordinates.
Here, you have A = (0, 0) and B = (3, 6). The horizontal distance between the points is ...
... 3 - 0 = 3 . . . . the difference of x-coordinates
The vertical distance between the points is ...
... 6 - 0 = 6 . . . . the difference of y-coordinates
Then the straight-line distance (d) between the points is found from the Pythagorean theorem, which tells you ...
... d² = 3² + 6²
... d = √(9 + 36) = √45 ≈ 6.7 . . . units
<span>The parent cosine function can be transformed and translated. So, from the basic function cos(x) we can obtain function acos(bx+c). In our case, a=3- amplitude, b=10- the period change and c=-pi- the phase shift. So, the parent cosine function is mutiplied with 3 (which gives the amplitude of the function, 3*0.5=1.5). The period of the function is changed, and is 2pi/b=2pi/10=pi/5 and the cos(x) is phase shifted for c/b=-pi/10.</span>
When solving system equations, we can use substitution method or elimination. Today I'm using substitution method.
First name the 2 equations.
3x + y = 3 (1)
x + y = 2 (2)
Now pick one equation and express one algebra in forms of the other.
From (2),
x = 2 - y (3)
Now substitute (3) into (1),
3(2-y) + y = 3
6 - 3y + y = 3
6 - 2y = 3
6 - 3 = 2y
y = 1.5
Now substitute y = 1.5 into (2)
x + 1. 5 = 2
x = 2 - 1.5
x = 0.5
Therefore the answer is x = 0.5 and y = 1.5
