A variable y is partly constant and partly varies as the square of x.When x =2,y=b and when x=3,y=10
a.find the equation connecting x and y
b.what is the value of when x=4
c.find the value of x when y =5
y = a + bx^2
Using the given values,
a+4b = 6 ??
a+9b = 10
Solve for a and b, and then you can answer the questions
Answer
a. Well, it seems like we have a case of a partially constant and partially clownishly squaring relationship between x and y. Let's solve this mystery together, shall we?
Given that y is partly constant, we can say that the constant part is b. And since y also varies as the square of x, we can represent this as:
y = b + kx^2, where k is the coefficient of the clownishly squaring relationship.
Now, let's substitute the given values to find b and k.
When x = 2, y = b:
b = b + k(2^2)
b = b + 4k
When x = 3, y = 10:
10 = b + k(3^2)
10 = b + 9k
Now, we have two equations:
b = b + 4k
10 = b + 9k
Well, this is getting a bit puzzling. Clearly, b is not going anywhere. So, to solve this conundrum, let's subtract the first equation from the second:
10 - (b = b + 4k)
10 - b = b - b + 9k - 4k
10 - b = 9k - 4k
10 - b = 5k
Now we know that 10 - b = 5k, and we can solve for k!
b. Now, let's put our detective skills to the test and find the value of y when x = 4:
Using the equation we found in the previous step, we have:
10 - b = 5k
Since we haven't figured out the values of b and k yet, we're trudging through a comically-ambiguous scenario, and it's best to say that we don't have enough information to find the value of y when x = 4. How mysterious!
c. With the information provided, we unfortunately can't directly find the value of x when y = 5. This clownish relationship has left us with a squinted eye, wondering about the secrets it holds. Maybe next time, we'll have more clues to solve this riddle.
To find the equation connecting x and y, we'll need to determine the constant and the quadratic term in the equation.
a. Let's start by assuming the equation connecting x and y is of the form y = c + ax^2. Since y is partly constant and partly varies as the square of x, the equation becomes y = c + ax^2.
To find the values of c and a, we'll use the given information. When x = 2, y = b, so we substitute these values into the equation: b = c + a(2^2) = c + 4a.
Similarly, when x = 3, y = 10. Substituting these values into the equation: 10 = c + a(3^2) = c + 9a.
We now have a system of equations:
b = c + 4a
10 = c + 9a
To solve for c and a, we'll subtract the first equation from the second equation:
10 - b = (c + 9a) - (c + 4a)
10 - b = 9a - 4a
10 - b = 5a
Now, divide both sides by 5 to isolate a:
a = (10 - b) / 5
Substituting this value of a into the first equation:
b = c + 4((10 - b) / 5)
b = c + 4(10 - b) / 5
b = c + (40 - 4b) / 5
5b = 5c + 40 - 4b
9b = 5c + 40
Therefore, the equation connecting x and y is y = (5c + 40) / 9 + ((10 - (5c + 40) / 9) / 5) * x^2.
b. To find the value of y when x = 4, we substitute x = 4 into the equation and solve for y:
y = (5c + 40) / 9 + ((10 - (5c + 40) / 9) / 5) * 4^2
y = (5c + 40) / 9 + ((10 - (5c + 40) / 9) / 5) * 16
c. To find the value of x when y = 5, we substitute y = 5 into the equation and solve for x:
5 = (5c + 40) / 9 + ((10 - (5c + 40) / 9) / 5) * x^2
5 - (5c + 40) / 9 = ((10 - (5c + 40) / 9) / 5) * x^2
By solving this equation, we can find the value of x when y = 5.
To find the equation connecting x and y, we need to determine the constant part and the part that varies as the square of x.
a. Finding the constant and varying parts:
When x = 2, y = b (given)
When x = 3, y = 10 (given)
We can rewrite the equation as:
y = c + dx^2
Substituting the values of x and y in the equation:
b = c + (2)^2
10 = c + (3)^2
Simplifying these equations:
b = c + 4 ⟹ (1)
10 = c + 9 ⟹ (2)
Now, solve equations (1) and (2) simultaneously to find the values of c and d:
Subtracting equation (1) from equation (2):
10 - b = c + 9 - c
10 - b = 9
b = 1
Substituting the value of b in equation (1):
1 = c + 4
c = -3
Therefore, the equation connecting x and y is:
y = -3 + dx^2
b. To find the value of y when x = 4, we need to substitute x = 4 in the equation we derived in part a:
y = -3 + d(4)^2
y = -3 + 16d
Since we do not have the value of d, we cannot determine the specific value of y when x = 4 without knowing the value of d.
c. To find the value of x when y = 5, we can rearrange the equation from part a:
y = -3 + dx^2
5 = -3 + dx^2
8 = dx^2
Simplifying further:
dx^2 = 8
x^2 = 8/d
To find the value of x, we need to know the specific value of d, which is not provided in the given information. Without the value of d, we cannot determine the specific value of x when y = 5.