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Sonbull [250]
2 years ago
6

Could the equation. For the graph of the function have degree 4?explain

Mathematics
1 answer:
aivan3 [116]2 years ago
5 0

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

#SPJ1

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3

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Then to resolve a quadratic equation we apply Bhaskara's formula:

x_{1}=\frac{-b+\sqrt{b^2-4ac} }{2a}

x_{2}=\frac{-b-\sqrt{b^2-4ac} }{2a}

It usually has two solutions.

But it could happen that \sqrt{b^2-4ac} then the equation doesn't have real solutions.

Or it could happen that there's only one solution, this happen when the linear equation touches the quadratic equation in one point.

And it's not possible to have more than 2 solutions. Then the answer ir 3.

For example:

In the three graphs the pink one is a quadratic function and the green one is a linear function.

In the first graph we can see that the linear function intersects the quadratic function in two points, then there are two solutions.

In the second graph we can see that the linear function intersects the quadratic function in only one point, then there is one solutions.

In the third graph we can see that the linear function doesn't intersect the quadratic function, then there aren't real solutions.

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Answer:

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Step-by-step explanation:

We have been given the face of the clock that is 63\ in

So that is also the circumference of the clock.

Since the clock is circular in shape.

So 2\pi(r)=63\ inch

From here we will calculate the value of radius (r) of the clock that is circular in shape.

Then 2\pi(r)=63\ inch =\frac{63}{2\pi} = 10.02\ in

Now to find the area of the clock we will put this value of (r) in the equation of area of the circle.

Now \pi (r)^{2}=\pi(10.02)^{2}=315.41\ in^{2}

So the area of the face of the clock =315.41\ in^{2}

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