A local zoo has flamingos and elephants in one exhibit. There are 16 animals in the exhibit, and the animals have a risk of 52 legs. How many of each type of animal are in the exhibit?

We have 6 flamy and 10 Elephants

Step-by-step explanation:

A local zoo has flamingos and elephants in one exhibit.

Flamingos have 2 legs

Elephants have 4 legs

Let the number of

Flamingos = x

Elephants = y

There are 16 animals in the exhibit

= x + y = 16 Equation 1

x = 16 - y

The animals have a risk of 52 legs.

2 × x + 4 × y = 52

2x + 4y = 52 Equation 2

We substitute 16 - y for x in Equation 2

2(16 - y) + 4y = 52

32 - 2y + 4y = 52

- 2y + 4y = 52 - 32

2y = 20

y = 20/2

y = 10

Solving for x

x = 16 - y

x = 16 - 10

x = 6

Hence, we have

Flamingos = x = 6 flamingos

Elephants = y = 10 Elephants

your answer is d 2,500

step-by-step explanation: