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

b = 6 when a=12 and

c=100

Step-by-step 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

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step-by-step explanation:

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .