First question: So, we know a fence that is 3 sections long, requires 4 posts and 6 rails.
Now, we have to use this kind of fence (3 sections, 4 posts, 6 rails) to fence off a yard. The yard is rectangular, and is 6 sections long and 3 sections wide.
Since we know the fence we are using is 3 sections wide, we have to count how many of them we’ll be using for the yard.
We know all four sides of the rectangle - two sides are 3 sections long, and the other two are 6 sections long.
Our fence is 3 sections, so we can already use 2 of those for the two sides of 3 sections. We multiply the number of rails and posts by two (since we are using two of these fences) and get 8 posts and 12 rails. We’ll keep this in mind.
Now, for the other two sides: they are both 6 sections long. We know that 3+3=6, thus one of those sides would be 2 fences long, or 3 sections times 2. We have two of these 6 section sides, so we actually would be using 12 sections (6*2), or in other words, we’d use this fence 4 times (4*3). We multiply the rails and posts by 4, and get 16 posts and 24 rails.
We use the 8 posts and 12 rails from earlier and add the number of the posts and rails we got just now: 16p and 24r, and get a total of 24 posts and 36 rails.
As for the second question, even I’m having a tough time. I’ll get back to you if I can solve it.
Answer:
C.) $1,035
Step-by-step explanation:
Edge 2020
Answer:
ah...yea-....want that to happen? ;P
jkjk...lol
Answer:
4 (20) + 6(6) = 120
4(20) + 4 (6) = 100
Step-by-step explanation:
Hi, to answer this question we have to write a system of equations with the information given:
Build time (hours) = 4c + 6a = 120
Test time (hours) = 4c + 4 a = 100
Where c represents the number of child bikes and a represents adult bikes.
So, for 20 child bikes and 6 adult bikes:
4 (20) + 6(6) = 120
4(20) + 4 (6) = 100
Feel free to ask for more if needed or if you did not understand something.
Answer:
A) Hertz: 42x+32=y
Enterprise: y=46x+24
B) Hertz is less by 16 $; 300$ minus 386$
C)16$
Step-by-step explanation:
