y is partly constant and partly varies as the cube of x given that whenx=1,y=5,andx=2,y=-1 find the relationship betweenx and y hence find the value of y when x=3
Do you mean that
y = a x^3 + b ?
If not, I don't see how y can be "partly constant" etc.
If so, then use your two known (x,y) points to establish a and b.
5 = a + b
-1 = 8a + b
7a = -6
a = -6/7
b = 41/7
y = -6x/7 + 41/7
Well, it seems like y is confused about whether it wants to be a constant or a variable - talk about an identity crisis! But fear not, I shall help you make sense of it.
If y is partly constant, that means it doesn't change when x changes. On the other hand, if it varies as the cube of x, then it does change with x. So, we have a bit of a contradiction here.
However, we can still find the relationship between x and y by interpreting the given information. We have two data points: when x = 1, y = 5, and when x = 2, y = -1. Let's plug these values into the equation.
When x = 1, y = k + 1^3 (k is the constant part of y). Plugging in the values, we get:
5 = k + 1
When x = 2, y = k + 2^3:
-1 = k + 8
Solving these two equations simultaneously, we can find the value of k:
k + 1 = 5
k = 4
So, the relationship between x and y is y = 4 + x^3.
Now, to find the value of y when x = 3, we can simply substitute x = 3 into the equation:
y = 4 + 3^3
y = 4 + 27
y = 31
Therefore, when x = 3, y = 31. Voila!
To find the relationship between x and y, we will first need to determine the constant part and the varying part of y.
Given that y is partly constant and partly varies as the cube of x, we can express y as follows:
y = c + kx^3
where c represents the constant part of y, k represents the coefficient of variation, and x^3 represents the varying part.
Using the given information that when x = 1, y = 5, we can substitute these values into the equation to obtain:
5 = c + k(1)^3
5 = c + k
Similarly, using the given information that when x = 2, y = -1:
-1 = c + k(2)^3
-1 = c + 8k
Now, we can solve these two equations simultaneously to find the values of c and k.
First, subtract the second equation from the first equation:
5 - (-1) = c + k - c - 8k
6 = -7k
Now, divide both sides of the equation by -7:
k = -6/7
Substituting this value of k into the first equation:
5 = c - 6/7
Multiplying both sides by 7 to remove the fraction:
35 = 7c - 6
7c = 41
c = 41/7
Now that we have the values of c and k, we can express y as a function of x:
y = (41/7) - (6/7)x^3
Finally, we can find the value of y when x = 3 by substituting x = 3 into the equation:
y = (41/7) - (6/7)(3)^3
y = 41/7 - (6/7)(27)
y = 41/7 - 162/7
y = (41 - 162)/7
y = -121/7
Therefore, when x = 3, y equals -121/7.
To understand the relationship between x and y, let's analyze the given information. We are told that y is partly constant and partly varies as the cube of x.
When x = 1, y = 5.
When x = 2, y = -1.
Let's break down the relationship between x and y into its two components: the constant part and the variable part.
1. Constant Part: The constant part of y indicates a value that does not change with x. In this case, it is not explicitly given, but we can determine it by substituting the known values. Since y = 5 when x = 1, the constant part of y is 5.
2. Variable Part: The variable part of y refers to the part that varies as the cube of x. We can determine the variable part by subtracting the constant part from the total value. So, in this case, the variable part is y - constant part = y - 5.
Now, let's substitute the known values to find the relationship between x and the variable part of y (y - 5).
When x = 1: y - 5 = 5 - 5 = 0.
When x = 2: y - 5 = -1 - 5 = -6.
We can observe that the value of the variable part follows the relationship y - 5 = k * x^3, where k represents a constant.
When x = 1: 0 = k * 1^3 = k.
When x = 2: -6 = k * 2^3 = 8k.
Now, finding the value of k:
k = 0, as determined when x = 1.
Now, we can express the relationship between x and y as y = k * x^3 + 5, where k = 0.
Therefore, the relationship between x and y is given by y = 0 * x^3 + 5, which simplifies to y = 5.
To find the value of y when x = 3:
y = 0 * 3^3 + 5 = 0 + 5 = 5.
Hence, when x = 3, y = 5.