★ Arithmetic Progression [ Explicit and Recursive notation ] ★
Arithmetic Sequence - 19 , 9 , -1 ...
Recursive Notation -
First term = 19
Common difference = -10
In subscript notation - [ a₁ = 19 , aₙ = aₙ-₁ -10 ]
In functional notation - f(n) = f (n -1) -10
Explicit Notation - f ( n ) = f ( 1 ) + d ( n - 1 )
= 19 - 10 ( n -1 )
= 29 - 10n
It's functional and subscript notations resides the same
To find the amount of fencing needed, we will find tge perimeter
To find the perimeter;
Find the distance |AB|, |BC|, |CD| and |DA|
We will use the formula below to find the distances:
![|d|=\sqrt[]{(x_2-x_1)^2+(y_{2-}y_1)^2}](https://tex.z-dn.net/?f=%7Cd%7C%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_%7B2-%7Dy_1%29%5E2%7D)
Distance |AB|
A(7,7), B(16,7)
x₁=7 y₁=7 x₂=16 y₂=7
Substituting into the formula;
![|AB|=\sqrt[]{(16-7)^2+(7-7)^2}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%5B%5D%7B%2816-7%29%5E2%2B%287-7%29%5E2%7D)
![=\sqrt[]{9^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B9%5E2%2B0%7D)
![=\sqrt[]{9}^2\text{ =9}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B9%7D%5E2%5Ctext%7B%20%3D9%7D)
Distance |BC|
B(16,7), C(2,2)
x₁=16 y₁=7 x₂=2 y₂=2
substituting into the formula;
![|BC|=\sqrt[]{(2-16)^2+(2-7)^2}](https://tex.z-dn.net/?f=%7CBC%7C%3D%5Csqrt%5B%5D%7B%282-16%29%5E2%2B%282-7%29%5E2%7D)
![=\sqrt[]{(-14)^2+(5)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-14%29%5E2%2B%285%29%5E2%7D)
![=\sqrt[]{196+25}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B196%2B25%7D)
![=\sqrt[]{221}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B221%7D)

Distance |CD|
C(2,2), D(16,2)
x₁=2 y₁=2 x₂=16 y₂=2
substituting into the formula;
![|CD|=\sqrt[]{(16-2)^2+(2-2)^2}](https://tex.z-dn.net/?f=%7CCD%7C%3D%5Csqrt%5B%5D%7B%2816-2%29%5E2%2B%282-2%29%5E2%7D)
![=\sqrt[]{14^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%2B0%7D)
![=\sqrt[]{14^2}=14](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%7D%3D14)
Distance |DA|
D(16,2) A(7,7)
x₁=16 y₁=2 x₂=7 y₂=7
Substituting into the formula;
![|DA|=\sqrt[]{(7-16)^2+(7-2)^2}](https://tex.z-dn.net/?f=%7CDA%7C%3D%5Csqrt%5B%5D%7B%287-16%29%5E2%2B%287-2%29%5E2%7D)
![=\sqrt[]{(-9)^2+(5)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-9%29%5E2%2B%285%29%5E2%7D)
![=\sqrt[]{81+25}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B81%2B25%7D)
![=\sqrt[]{106}=10.30](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B106%7D%3D10.30)
Perimeter = |AB|+|BC|+|CD|+|DA|
= 9 + 14.87 + 14 + 10.30
=48.17
≈48
Hence;
48 feet of fencing is needed
First make the denomonators the same
Answer:
x = 10.6
y = 10.6
Step-by-step explanation:
To answer this question you will need to use trigonometry. If you don't already have an understanding of trigonometry, this will be very hard to explain. please let me know in the comments if you would like me to give it a go.
I hope this was helpful :-)
If 1 ml = 0.034 ounces
ounces in 1 hectoliter = ?
1ml = 0.001 litre
1 liter = 1000ml
1 hectoliter = 100000ml
1 hectolitre = 100000 x 0.034 ounces
1 hectolitre = 3400 ounces
So, 3400 ounces are in 1 hectoliter of fluid.