M is partly constant and partly varies as n square when n=2,m=17 and when n=2,m=53 find (a)the equation conneting m and n (b)n when m=32

32×4=

m = a + bn^2

so, using the two data points,
a+4b=17
a+4b=53
How can m be both 17 and 53 when n=2? Fix that and then solve for a and b, and then you can find n when m=32

(a) Well, let's break some ice and start with a little math humor. The equation connecting m and n would be as intriguing as a clown at a magic show. So, since we know that M is partly constant and partly varies with n squared, we can express it as:

M = k + a * n^2

Where k is the constant part and a is the coefficient of the varying part. Now, let's substitute the given values:

17 = k + a * 2^2
53 = k + a * 2^2

Solving these equations simultaneously, we find that k = 9 and a = 4. Therefore, the equation connecting m and n is:

M = 9 + 4 * n^2

(b) Let's put on our clown shoes and find some laughter in math. We're looking for n when m equals 32. So, we can plug that into our equation:

32 = 9 + 4 * n^2

By rearranging the equation, we have:

4 * n^2 = 32 - 9
n^2 = 23/4

Taking the square root of both sides, we get:

n = √(23/4)

And since we're all about jokes here, we'll simplify it a bit:

n ≈ √(5.75)

So, when m equals 32, n is approximately equal to the square root of 5.75. Keep smiling, my friend!

To find the equation connecting m and n, we need to find the relationship between the constant part and the variable part.

We are given two points where n=2 and m=17, and n=2 and m=53. Let's denote the constant part as k and the variable part as n^2.

Using the given points:
For n=2, m=17: m = k + (n^2) = k + (2^2) = k + 4
For n=2, m=53: m = k + (n^2) = k + (2^2) = k + 4

Since both equations have the same constant, k + 4 = 17 and k + 4 = 53, we can solve for k.

k + 4 = 17
k = 17 - 4
k = 13

Now we can substitute the value of k into the equation and find the equation connecting m and n:
m = 13 + (n^2)

To find n when m=32, we can substitute m=32 into the equation and solve for n:
32 = 13 + (n^2)

Rearranging the equation to isolate n^2:
(n^2) = 32 - 13
(n^2) = 19

Taking the square root of both sides, we get:
n = ±sqrt(19)

So, when m=32, n can be either n = sqrt(19) or n = -sqrt(19).

To find the equation connecting m and n when m is partly constant and partly varies as n squared, we can utilize the given information.

Let's break down the information provided:

When n = 2, m = 17
When n = 2, m = 53

We are given two points on the graph, which will help us form the equation.

(a) Finding the equation connecting m and n:
To determine the relationship between m and n, we need to identify the constant part and the varying part.

From the given information, we can see that when n = 2, m is constant at 17. This means that the constant part of m is 17.

Now, let's look at how m varies when n = 2. We can calculate the difference between the two given m values:

53 - 17 = 36

We know that the variation is proportional to n squared, so let's find the square of the difference in n values:

(2^2) - (2^2) = 4 - 4 = 0

The constant part of m is 17, and the varying part is 36 times the square of n (which in this case is 2).

Therefore, the equation connecting m and n is:
m = 17 + 36n^2

(b) Finding n when m = 32:
To find n when m = 32, we can substitute m = 32 into the equation and solve for n.

32 = 17 + 36n^2

Rearranging the equation, we get:

36n^2 = 32 - 17
36n^2 = 15

Divide both sides of the equation by 36:

n^2 = 15/36
n^2 = 5/12

To solve for n, we take the square root of both sides:

n = ±√(5/12)

So, n can be either the positive or negative square root of 5/12.