The graphed polynomial seems to have a degree of 2, so the degree can be 4 and not 5.
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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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First: the homogeneous solutions: the characteristic equation is4r^2 - 4r - 3 = 0which has roots r = 3/2, -1/2 hence the homogeneous solution isy = c1.exp(-x/2) + c2.exp(3x/2)
next you need the general form for the guess for yp and that isyp = A1cos(2x) + A2sin(2x)
Now substitute that into the equation and solve for A1, A2.
When (-5, 2) is reflected across the y-axis then it becomes (5, 2). When (-5, 2) is reflected across the x-axis, it becomes (-5, -2)
I hope this helps out!
Answer:
The fraction converted into percentage is .68.96551724%
Answer:
105 ways
Step-by-step explanation:
Because the order of selection does not matter, we know that we will be using a combination instead of a permutation.
₁₅C₂ = 15! / (2! * 13!)
= 15 * 14 * ... * 2 * 1 / 2 * 1 * 13 * 12 * ... * 2 * 1
= 15 * 14 / 2 * 1 (The 13! cancels out)
= 105
