There are 46 flags in a zoo exhibit. Every animal has either two or four legs.
List 2 possible choices for the number of each type of animal in the exhibit.
The number of two-legged animals is two more than a number of four-legged animals. How many of each are the exhibit?
46 flags????
You purchase four videos. The original price of each video is X dollars. You decide to purchase the limited edition versions of the videos for additional cost. Total cost is (4x+20) dollars. What can you conclude about the additional cost of the limited edition version of the video?
Please do not change names.
You haven't answered my question about your first post.
There are 46 legs in a zoo exhibit. Every animal has either two or four legs.
List 2 possible choices for the number of each type of animal in the exhibit.
The number of two-legged animals is two more than a number of four-legged animals. How many of each are the exhibit?
4x + 2(x+2) = 46
Solve for x.
To solve this problem, let's use algebra to represent the number of two-legged animals and four-legged animals in the exhibit.
Let's assume the number of two-legged animals is "x" and the number of four-legged animals is "y".
According to the information given, there are a total of 46 flags in the zoo exhibit, which represents the total number of animals. So we can write the equation:
x + y = 46
Now, we are also given that the number of two-legged animals is two more than the number of four-legged animals. So we can write another equation:
x = y + 2
Now we have a system of two equations:
x + y = 46
x = y + 2
We can solve this system of equations to find the values of x and y.
First, let's substitute the value of x from the second equation into the first equation:
(y + 2) + y = 46
Combining like terms, we get:
2y + 2 = 46
Subtracting 2 from both sides, we get:
2y = 44
Now divide both sides by 2:
y = 22
Now substitute the value of y back into the second equation to find x:
x = 22 + 2 = 24
So, there are 24 two-legged animals and 22 four-legged animals in the exhibit.