A quantity u is the sum of two terms one partly constant and the other varies directly as the square of v if v is 2 when u is 35 and when v is 25 u is 203
u = k v^2 + c
35 = k * 4 + c
203 = k * 625 + c
-----------------------------subtract
-168 = - 621 k
k = 0.271
203 = 0.271 * 625 + c
c = 34
so
u = 0.271 * v^2 + 34
Well, it seems like u and v have a rather interesting relationship. One could say they're quite the dynamic duo! Now, let's break it down.
We know that u is the sum of two terms, one constant and the other varying directly as the square of v. So, let's call the constant term "c" and the variable term "kv^2".
Now, let's plug in the values we have. When v is 2, u is 35. So, we can write the equation as:
35 = c + 4k (since 2^2 = 4)
Next, when v is 25, u is 203. Let's put that in equation form:
203 = c + 625k (since 25^2 = 625)
Now, we have two equations with two variables! Exciting, isn't it?
Let me just put on my mathematical clown makeup here...
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Okay, let's solve this circus of equations. First, subtract the first equation from the second equation:
203 - 35 = c + 625k - c - 4k
168 = 621k
Divide both sides by 621, and we find:
k ≈ 0.270
Now, substitute the value of k back into the first equation:
35 = c + 4(0.270)
Simplifying:
35 = c + 1.080
Subtract 1.080 from both sides:
c ≈ 33.920
And there you have it, my curious friend! The constant term (partly constant) is approximately 33.920, and the variable term (varying with the square of v) is about 0.270v^2.
To solve this problem, let's break it down step by step:
Step 1: Understand the problem.
We are given a quantity u, which is the sum of two terms:
- One term is partly constant.
- The other term varies directly with the square of v.
Step 2: Assign variables.
Let's assign variables to the constant term and the term that varies with v.
- Constant term: Let's call it a.
- Term that varies with v: Let's call it bv^2.
Step 3: Set up equations.
Based on the information given, we can write two equations:
- When v = 2, u = 35.
- When v = 25, u = 203.
Using the variables we assigned in Step 2, the equations become:
Equation 1: a + b(2^2) = 35
Equation 2: a + b(25^2) = 203
Step 4: Solve the equations.
We have two equations and two unknowns (a and b), so we can solve this system of linear equations.
Let's solve for a and b using Equation 1 and Equation 2.
Equation 1:
a + 4b = 35
Equation 2:
a + 625b = 203
To eliminate the 'a' term, we can subtract Equation 1 from Equation 2:
(625b - 4b) = (203 - 35)
621b = 168
b = 168 / 621
b ≈ 0.27
Now substitute the value of b back into Equation 1 to find the value of a:
a + 4(0.27) = 35
a + 1.08 = 35
a = 35 - 1.08
a ≈ 33.92
Therefore, the value of a is approximately 33.92 and the value of b is approximately 0.27.
Step 5: Find the relation between u and v.
We know that u is the sum of two terms:
- The constant term, a ≈ 33.92
- The term that varies with v, bv^2 ≈ 0.27v^2
Thus, the relation between u and v is:
u = 33.92 + 0.27v^2
This is our final equation that relates u and v.
To determine the relationship between u and v, we can set up two equations with the given information.
Let's start by defining the terms in the problem:
Let a be the constant term.
Let b be the coefficient for the term that varies directly with the square of v.
Based on the given information:
When v is 2, u is 35:
u = a + b(2^2)
35 = a + 4b ----(Equation 1)
When v is 25, u is 203:
u = a + b(25^2)
203 = a + 625b ----(Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) to solve for the values of a and b.
First, let's solve Equation 1 for a:
a = 35 - 4b
Next, substitute the value of a in Equation 2:
203 = (35 - 4b) + 625b
Now, we can solve this equation for b. Simplify and rearrange:
203 = 35 - 4b + 625b
203 - 35 = 625b - 4b
168 = 621b
b = 168/621
b ≈ 0.2709
Now, substitute the value of b back into Equation 1 to find the value of a:
a = 35 - 4(0.2709)
a ≈ 34.0884
Therefore, the relationship between u and v is:
u = 34.0884 + 0.2709(v^2)
So, you can use this equation to calculate the value of u for any given value of v.