Below are two regular polygons. Calculate the value of angle a.
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Answer:
See attached image for the drawing of the first four trees (circled in green)
The patterns is:
x = 2+3n and y= 3+2n
Position of 7th tree is: (20,15) (circled in orange in the image)
Step-by-step explanation:
Starting at the location (2,3) the next x and y positions are given by:
x = 2+3n since the horizontal position needs to be increased by 3 units on each iteration,
and y= 3+2n since the vertical position needs to be increased by 2 units on each iteration
being n= 1 through 6 (to account for the next 6 trees that need to be planted)
With such pattern, the location of the seventh tree would be:
x = 2 + 3*6 =2 + 18 = 20
y = 3 + 2*6 = 3 + 12 = 15
That is, the point (20,15) on the plane.
Also see attached image.
Check the picture below.
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,
![\bf \begin{cases} h=2\\ k=-3\\ a=6\\ c=2\sqrt{10} \end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2 \\\\\\ 4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2 \\\\\\ \sqrt{4}=b\implies 2=b\\\\ -------------------------------\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ah%3D2%5C%5C%0Ak%3D-3%5C%5C%0Aa%3D6%5C%5C%0Ac%3D2%5Csqrt%7B10%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5By-%20%28-3%29%5D%5E2%7D%7B%206%5E2%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%282%5Csqrt%7B10%7D%29%5E2%3D6%5E2%2Bb%5E2%5Cimplies%202%5E2%28%5Csqrt%7B10%7D%29%5E2%3D36%2Bb%5E2%0A%5C%5C%5C%5C%5C%5C%0A4%2810%29%3D36%2Bb%5E2%5Cimplies%2040%3D36%2Bb%5E2%5Cimplies%204%3Db%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%7B4%7D%3Db%5Cimplies%202%3Db%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%202%5E2%7D%3D1%5Cimplies%20%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%204%7D%3D1)
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,