Answer:
![Area = 93.21in^2](https://tex.z-dn.net/?f=Area%20%3D%2093.21in%5E2)
Step-by-step explanation:
Given
vertex angle
base angle
![\theta = 3\alpha](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%5Calpha)
-- congruent sides
Required
The area of the triangle
First, calculate the angles
----- angles in an isosceles triangle
Substitute ![\theta = 3\alpha](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%5Calpha)
![3\alpha + \alpha + \alpha = 180^o](https://tex.z-dn.net/?f=3%5Calpha%20%2B%20%5Calpha%20%2B%20%5Calpha%20%3D%20180%5Eo)
![5\alpha = 180^o](https://tex.z-dn.net/?f=5%5Calpha%20%3D%20180%5Eo)
Divide both sides by 5
![\alpha = 36^o](https://tex.z-dn.net/?f=%5Calpha%20%3D%2036%5Eo)
Recall that: ![\theta = 3\alpha](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%5Calpha)
![\theta = 3 * 36^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%20%2A%2036%5Eo)
![\theta = 108^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20108%5Eo)
The area is then calculated as:
![Area = \frac{1}{2}l^2 \sin(\theta)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7Dl%5E2%20%5Csin%28%5Ctheta%29)
![Area = \frac{1}{2}*14^2 \sin(108^o)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A14%5E2%20%5Csin%28108%5Eo%29)
![Area = \frac{1}{2}*196 *0.9511](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A196%20%2A0.9511)
![Area = 93.21in^2](https://tex.z-dn.net/?f=Area%20%3D%2093.21in%5E2)
Answer:
FOR H_2
![1 \mu mol/L = 2\times 10^{-6} mol/L](https://tex.z-dn.net/?f=1%20%5Cmu%20mol%2FL%20%3D%202%5Ctimes%2010%5E%7B-6%7D%20mol%2FL)
Step-by-step explanation:
Given data:
concentration ![0.6 \mu mol/L](https://tex.z-dn.net/?f=0.6%20%5Cmu%20mol%2FL)
WE KNOW THAT
![ONE\ \mu = 10^{-6}](https://tex.z-dn.net/?f=ONE%5C%20%20%5Cmu%20%3D%2010%5E%7B-6%7D)
therefore
![1 \mu mol/L = 10^{-6} mol/L](https://tex.z-dn.net/?f=1%20%5Cmu%20mol%2FL%20%3D%2010%5E%7B-6%7D%20mol%2FL)
Now for conversion of mol to gram need the molar mass of element in which particular value convert to.
let us take ![H_2](https://tex.z-dn.net/?f=H_2)
we know that molar mass of
is 2 gram
therefore
![1 \mu mol/L = 2\times 10^{-6} mol/L](https://tex.z-dn.net/?f=1%20%5Cmu%20mol%2FL%20%3D%202%5Ctimes%2010%5E%7B-6%7D%20mol%2FL)
let take other example, for ![O_2](https://tex.z-dn.net/?f=O_2)
we know that molar mass of O_2 is 32 gram
therefore
![1 \mu mol/L = 32\times 10^{-6} mol/L](https://tex.z-dn.net/?f=1%20%5Cmu%20mol%2FL%20%3D%2032%5Ctimes%2010%5E%7B-6%7D%20mol%2FL)
Answer:
1,200
Steps down below
Step-by-step explanation:
Steps
1: 48=4% loaded on truck and they have 96% left to load
2: so set up a proportion and it should look like this
3: 48 over x = 4 over 100
4: 48/x=4/100
6: do cross multiplication 100×48= 4,800÷ 4 = 1,200 total on truck
Answer: 1,200 total on truck
<em><u>Hope this helps.</u></em>