if we plot those 4 points, the two vertices and the two foci, we'll notice that they're all colinear vertically, namely making a vertical line, which in short means is a vertical ellipse, which means the "a" component will go under the fraction with the "y" above.
the half-way between the vertices lands on (-2, -3), thus that is our center.
the distance from one vertex to the other is 10 units, meaning a = 5.
the distance from the center to either foci is 4 units, meaning c = 4.
what's "b" anyway?
![\bf \textit{ellipse, vertical major axis} \\\\ \cfrac{(x- h)^2}{ b^2}+\cfrac{(y- k)^2}{ a^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h = -2\\ k = -3\\ a = 5\\ c = 4 \end{cases}\implies \cfrac{[x-(-2)]^2}{b^2}+\cfrac{[y-(-3)]^2}{5^2}=1 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bellipse%2C%20vertical%20major%20axis%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20b%5E2%7D%2B%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20a%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%2C%20k%5Cpm%20a%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2-%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%20%3D%20-2%5C%5C%20k%20%3D%20-3%5C%5C%20a%20%3D%205%5C%5C%20c%20%3D%204%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Bx-%28-2%29%5D%5E2%7D%7Bb%5E2%7D%2B%5Ccfrac%7B%5By-%28-3%29%5D%5E2%7D%7B5%5E2%7D%3D1%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf c = \sqrt{a^2-b^2}\implies 4=\sqrt{5^2-b^2}\implies 4^2=5^2-b^2\implies \implies b^2+4^2=5^2 \\\\\\ b^2=5^2-4^2\implies b^2=25-16\implies b^2=9\implies b=\sqrt{9}\implies \boxed{b = 3} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[x-(-2)]^2}{3^2}+\cfrac{[y-(-3)]^2}{5^2}=1\implies \cfrac{(x+2)^2}{9}+\cfrac{(y+3)^2}{25}=1](https://tex.z-dn.net/?f=%5Cbf%20c%20%3D%20%5Csqrt%7Ba%5E2-b%5E2%7D%5Cimplies%204%3D%5Csqrt%7B5%5E2-b%5E2%7D%5Cimplies%204%5E2%3D5%5E2-b%5E2%5Cimplies%20%5Cimplies%20b%5E2%2B4%5E2%3D5%5E2%20%5C%5C%5C%5C%5C%5C%20b%5E2%3D5%5E2-4%5E2%5Cimplies%20b%5E2%3D25-16%5Cimplies%20b%5E2%3D9%5Cimplies%20b%3D%5Csqrt%7B9%7D%5Cimplies%20%5Cboxed%7Bb%20%3D%203%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%5Bx-%28-2%29%5D%5E2%7D%7B3%5E2%7D%2B%5Ccfrac%7B%5By-%28-3%29%5D%5E2%7D%7B5%5E2%7D%3D1%5Cimplies%20%5Ccfrac%7B%28x%2B2%29%5E2%7D%7B9%7D%2B%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B25%7D%3D1)
Check the picture below.
(2/3)(2/3)(2/3)(2/3)
=16/81
Answer:
1 1/5 mi
Step-by-step explanation:
We need to add the three legs together
3/10 + 1/2 + 2/5
The common denominator is 10
3/10 =3/10
1/2 *5/5 = 5/10
2/5 *2/2 =4/10
3/10 + 5/10+4/10 = 12/10
10/10 = 10/10+2/10 = 1+2/10 = 1 1/5 mi
Answer:
Solution given:
perpendicular [P]=15m
base[B]=8m
hypotenuse [H]=17m
rate :4 plants per square metre
no of plants=?
we have
Area of triangular garden: ½(P*B)=½(15*8)=60m²
Now
Total no of plants =rate ×Area =4×60=240
Michael needs<u> 240</u> plants in the garden