Question

Suppose b varies directly with a and inversely with the square root of c. If b=5 when a=2 and c=4, find b when a=12 and

c=100
What if b

Answer 1

b = 6 when a=12 and

c=100

Explanation:

b = (k*a)/√c

b=5 when a=2 and c=4

5 = (k*2)/√4

5 = 2k/2

Cross product

5*2 = 2k

10 = 2k

k = 10/2

= 5

k = 5

find b when a=12 and

c=100

b = (k*a)/√c

= (5*12)/√100

= 60/10

= 6

b = 6