A is partly constant and partly varies as square of B when b=1 a=4¼ and when b=7 a=196¼, (a) find the equation connection a and b (b) find a when b=square root of 15/4
"A is partly constant and partly varies as square of B"
----> A = mB^2 + c
when b = 1, A = 4.25
4.25 = m + c
when b=7 a=196¼
196.25 = 49m + c
subtract
192 = 48m
m = 4
then 4.25 = 4 + c, c = 0.25
A = 4B^2 + .25
now you can state the actual equation, take it from here.
I don't know how to solve it
Teach me please
To find the equation connecting a and b, we can assume that the relationship between a and b is of the form:
a = k * B^2 + c
where k and c are constants that we need to determine.
Given that when b = 1, a = 4¼, we can substitute these values into our equation:
4¼ = k * 1^2 + c
4¼ = k + c ----(1)
Similarly, when b = 7, a = 196¼, we can substitute these values into our equation:
196¼ = k * 7^2 + c
196¼ = 49k + c ----(2)
Now we have a system of two equations (1) and (2) with two unknowns (k and c). We can solve this system of equations to find the values of k and c.
Subtracting equation (1) from equation (2):
196¼ - 4¼ = 49k + c - k - c
192 = 48k
Dividing both sides by 48:
4 = k
Substituting the value of k back into equation (1):
4¼ = 4 + c
4¼ - 4 = c
¼ = c
So, we have determined that k = 4 and c = ¼.
Therefore, the equation connecting a and b is:
a = 4B^2 + ¼
To find the value of a when b = √(15/4), we can substitute this value into the equation:
a = 4(√(15/4))^2 + ¼
a = 4(15/4) + ¼
a = 15 + ¼
a = 15¼
To find the equation connecting "a" and "b," we can assume that "a" is composed of two parts: a constant part, let's call it "c," and a variable part, let's call it "k * B^2", where "k" is a constant.
So, the equation connecting "a" and "b" can be written as:
a = c + k * B^2
Now, let's find the values of "c" and "k" using the given information:
When b = 1 and a = 4¼:
4¼ = c + k * 1^2 (substituting the values of a and b)
4¼ = c + k (since 1^2 = 1)
c + k = 4¼ (equation 1)
When b = 7 and a = 196¼:
196¼ = c + k * 7^2
196¼ = c + 49k (since 7^2 = 49)
c + 49k = 196¼ (equation 2)
Now, we have two equations with two unknowns (c and k). We can solve these equations simultaneously to determine their values.
Subtracting equation 1 from equation 2, we get:
( c + 49k ) - ( c + k ) = 196¼ - 4¼
48k = 192 (since c cancels out)
k = 4 (dividing both sides by 48)
Substituting the value of k in equation 1, we get:
c + 4 = 4¼
c = ¼
Therefore, the equation connecting "a" and "b" is:
a = ¼ + 4 * B^2
Now, let's find the value of "a" when b = √(15/4):
Substituting the value in the equation:
a = ¼ + 4 * (√(15/4))^2
a = ¼ + 4 * (15/4)
a = ¼ + 15
a = 15¼
So, when b = √(15/4), a = 15¼.