This line through M and parallel to BC is the <span>average line of BC. So this lane goes through the midpoint of AC that is N. And we have the length of AN=1/2AC=3/2</span>
Using pythagorean identity:
![\sin ^2(\theta)+\cos ^2(\theta)=1](https://tex.z-dn.net/?f=%5Csin%20%5E2%28%5Ctheta%29%2B%5Ccos%20%5E2%28%5Ctheta%29%3D1)
Divide both sides by cos²:
![\tan ^2(\theta)+1=\sec ^2(\theta)](https://tex.z-dn.net/?f=%5Ctan%20%5E2%28%5Ctheta%29%2B1%3D%5Csec%20%5E2%28%5Ctheta%29)
Replace the given values:
![\begin{gathered} (\frac{24}{7})^2+1=(\frac{25}{7})^2 \\ 1+(\frac{24}{7})^2=(\frac{25}{7})^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28%5Cfrac%7B24%7D%7B7%7D%29%5E2%2B1%3D%28%5Cfrac%7B25%7D%7B7%7D%29%5E2%20%5C%5C%201%2B%28%5Cfrac%7B24%7D%7B7%7D%29%5E2%3D%28%5Cfrac%7B25%7D%7B7%7D%29%5E2%20%5Cend%7Bgathered%7D)
Answer:
Answer:
Few animal combinations of 3 animals given income & prices :- Only one of cow, horse, or ship bought ; less than 100 , 200 , 200 respectively.
Step-by-step explanation:
Budget Set is the combination of goods, which can be afforded by a consumer, given money income & prices.
Given : Budget Income = 100
Cow Price (C) = 1 ; Sheep Cost (S) = 0.5 ; Horse Cost (H) = 0.5
Maximum number of either cow, or sheep, or horse only that Farmer can buy with given prices & budget income (not buying the other ones):-
- C = 100 / 1 = 100
- S = 100 / 0.5 = 200
- H = 100 / 0.5 = 200
So, some animal combination, that farmer can buy with available income budget & given prices :-
- ( C , 0 , 0 ) ; where C < 100
- ( 0 , S , 0 ) ; where S < 200
- ( 0 , 0, H ) ; where H < 200
The answer is 12 inches :D