If a cow costs 1$ a sheep costs 0.5$ and a horse cost 10$ and you have 100$ to buy 100 animals and you have to have one of each. how many of each did you buy?

5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.

Step-by-step explanation:

Cost of 1 cow = $1

Cost of 1 sheep = $0,5

Cost of 1 horse = $10

Let number of cows be x and , number sheep be y and number of horses be z.

...(1)

..(2)

Since , there should be minimum 1 animal. we will be using hit and trial method for the solution:

1) Put value of x =1 ,  in (1) and (2),we get two equation in two variables:

Solving above equation we get , y =93.79 , z= 5.21 (Not possible)

2) Put value of x = 2 ,  in (1) and (2),we get two equation in two variables:

Solving above equation we get , y =92.85 , z= 5.15 (Not possible)

3)  Put value of x = 5 ,  in (1) and (2), we get two equation in two variables:

Solving above equation we get , y =90 , z= 5 (possible)

5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.