A is partly constant and partly varies as b when A is equal to 15 B was 2 when is equal to A 45 is B equal to 8
A = n B + k
15 = 2 n + k
45 = 8 n + k
subtracting equations ... 30 = 6 n
solve for n , then substitute back to find k
To determine when B is equal to 8 when A is equal to 45, we can use the given information: "A is partly constant and partly varies as b."
Let's break down the problem step-by-step:
1. We know that when A is equal to 15, B was 2. This implies that a part of A is constant and equals 15, and the other part is proportional to B.
2. If A is partly constant and partly varies as b, we can represent it as:
A = p + q * B,
where p is the constant part and q is the proportionality constant.
3. We know that when A is 15, and B is 2, then we can substitute these values into the equation:
15 = p + q * 2.
4. Now, let's determine the values of p and q. We can rearrange the equation from step 3 to solve for either p or q. Let's rearrange it to solve for p:
p = 15 - q * 2.
5. With the equation from step 4, let's substitute A = 45 and solve for q:
45 = (15 - q * 2) + q * B.
Simplify the equation:
45 = 15 - q * 2 + q * 8.
Rearrange the equation to isolate q:
45 - 15 = q * 8 - q * 2.
30 = q * (8 - 2).
30 = q * 6.
Divide both sides by 6 to solve for q:
q = 30 / 6 = 5.
6. Now that we have found the value of q, we can substitute it back into the equation from step 4 to find p:
p = 15 - q * 2 = 15 - 5 * 2 = 15 - 10 = 5.
7. Therefore, when A is equal to 45, and B is 8, we have:
A = p + q * B = 5 + 5 * 8 = 5 + 40 = 45.
B = 8.
So when A is 45, B is also equal to 8.
To solve this problem, we need to find the relationship between A and B.
We are given that A is partly constant and partly varies as B. This means that there is a constant term (let's call it C) that is added to a term that is directly proportional to B.
Mathematically, we can express this relationship as:
A = C + K * B
where C is the constant term and K is the proportionality constant.
Now, we are given two specific points on this relationship: when A is equal to 15, B was 2, and when A is equal to 45, B is 8.
Let's use these points to find the values of C and K.
When A is equal to 15 and B is equal to 2, we can substitute these values into the relationship equation:
15 = C + K * 2
Similarly, when A is equal to 45, and B is equal to 8:
45 = C + K * 8
Now, we have a system of two equations:
15 = C + 2K
45 = C + 8K
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From the first equation, we can isolate C:
C = 15 - 2K
Substituting this value of C into the second equation:
45 = (15 - 2K) + 8K
Simplifying:
45 = 15 + 6K
30 = 6K
K = 5
Now, we can substitute this value of K back into the first equation to find C:
15 = C + 2 * 5
15 = C + 10
C = 5
So, the constant term C is equal to 5, and the proportionality constant K is equal to 5.
Finally, we can use these values to find B when A is equal to 45:
A = C + K * B
45 = 5 + 5 * B
Solving for B:
40 = 5B
B = 8
Therefore, when A is equal to 45, B is equal to 8.