Answer:
1) r, q, s. 2). 5
Step-by-step explanation:
1) from the given (q) is smaller than (r) and (s) is less than (q) by 2.
2) you need to make all the fractions have the same numerator so you will multiply both the firs and third fraction by 3÷3 ( which is 1 and will not affect the value but will make the numerator of the fractions equal) and the fractions between them is the second fraction and it's denominator is x and there is 5 fractions between the two numbers so there are 5 possible values of x
You multiply it with the same number you use for the denominator.
in this case 3/4 and 7/10
you find the LCM of the denominator is 20
so for 4 to become 20 you multiply by 5 then for the numerator 3 also multiply by 5, so you get 15/20
its the same also with 7/10, you multiply both the numerator and the denominator by 2 and get 14/20
hope it help... :)
If we take the number of Zebras to be x and that of peacocks to be y
Then x + y = 14
A zebra has 4 legs while a peacock has 2 legs,
Therefore the total number of legs will be 4x + 2y which is equal to 36 .
Therefore, we have two equations
x + y = 14
4x + 2y = 36 solving them simultaneously using elimination
4x +2y = 36
2x + 2y = 28 ( multiplying the first equation by 2 to eliminate y)
2x = 8 ( subtracting the equations)
x = 4
y = 14 -(4)
= 10
Therefore, there are 4 zebras and 10 peacocks in the zoo
