if A varies directly as the square root of B and A varies directly as the square of C,when A=6, C=3 find the equation connecting the A and C
a = kc^2
6 = 9k
a = 2/3 c^2
To find the equation connecting A and C, we need to consider the two given relationships:
1) A varies directly as the square root of B: A ∝ √B
2) A varies directly as the square of C: A ∝ C^2
Since A varies directly as the square root of B, we can express it as:
A = k√B (where k is the constant of proportionality)
Similarly, since A varies directly as the square of C, we can express it as:
A = mC^2 (where m is the constant of proportionality)
Given that A = 6 and C = 3, we can substitute these values into one of the equations to find the constant of proportionality:
A = k√B
6 = k√B
Next, let's substitute the value of C into the second equation:
A = mC^2
6 = m(3)^2
6 = 9m
Now we have two equations:
6 = k√B
6 = 9m
Squaring both sides of the first equation to eliminate the square root:
(6)^2 = (k√B)^2
36 = k^2B
Dividing both sides of the second equation by 9:
6/9 = 9m/9
2/3 = m
Now we have two equations:
36 = k^2B
2/3 = m
To find an equation connecting A and C, we need to eliminate the constants of proportionality, k and m.
Let's substitute the value of m into the second equation:
2/3 = m
2/3 = k√B
2/3 = (√B)/3
Multiplying both sides of the equation by 3/2 to isolate (√B):
(2/3) * (3/2) = (√B)/3 * (3/2)
1 = (√B)/2
Squaring both sides to eliminate the square root:
1^2 = ((√B)/2)^2
1 = B/4
Now we have an equation connecting A and C:
A = k√B
A = (√B)/2
Simplifying A = (√B)/2 by substituting B = 4:
A = (√4)/2
A = 2/2
A = 1
Therefore, the equation connecting A and C is:
A = 1 * C^2
A = C^2
To find the equation connecting A and C, we need to understand the given information.
We know that "A varies directly as the square root of B", which can be represented as:
A = k√B, where k is the constant of variation.
We also know that "A varies directly as the square of C", which can be represented as:
A = mC^2, where m is another constant of variation.
Since we are given that A = 6 when C = 3, we can substitute these values into the equation to determine the values of k and m.
From the first equation:
6 = k√B
From the second equation:
6 = m(3)^2
Simplifying the second equation:
6 = 9m
Solving for m:
m = 6/9 = 2/3
Substituting m = 2/3 into the first equation:
6 = k√B
Multiplying both sides by √B:
6√B = kB
Now, we need to solve for k. Since we have two equations involving k, we can equate them to each other:
6√B = kB
Dividing both sides by B:
6/√B = k
Since we have found the values of k and m, we can combine them to form the equation connecting A and C:
A = k√B = (6/√B)√B = 6
Now, let's substitute the value of m into the equation:
A = (2/3)C^2
Therefore, the equation connecting A and C is A = (2/3)C^2.