Simplified would be: 3s + 2
Answer:
Developed Length
Step-by-step explanation:
A developed length is the length along the centre line of a circular shape.
From the question, the degree of bend is 90°
Also, from the question; the length of an arc of 1 quadrant.
1 quadrant = 90°
And the degree of bend = 90° (from the question)
So, it's safe to say 1 quadrant = the degree of bend = 90°
A small change in 90° will move change the quadrant to another quadrant.
So, at exactly 90° (or other multiples of 90° which are 180°, 270° and 360°), it is at its developed length.
The graphed polynomial seems to have a degree of 2, so the degree can be 4 and not 5.
<h3>
Could the graphed function have a degree 4?</h3>
For a polynomial of degree N, we have (N - 1) changes of curvature.
This means that a quadratic function (degree 2) has only one change (like in the graph).
Then for a cubic function (degree 3) there are two, and so on.
So. a polynomial of degree 4 should have 3 changes. Naturally, if the coefficients of the powers 4 and 3 are really small, the function will behave like a quadratic for smaller values of x, but for larger values of x the terms of higher power will affect more, while here we only see that as x grows, the arms of the graph only go upwards (we don't know what happens after).
Then we can write:
y = a*x^4 + c*x^2 + d
That is a polynomial of degree 4, but if we choose x^2 = u
y = a*u^2 + c*u + d
So it is equivalent to a quadratic polynomial.
Then the graph can represent a function of degree 4 (but not 5, as we can't perform the same trick with an odd power).
If you want to learn more about polynomials:
brainly.com/question/4142886
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