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3 years ago

Whats it called when 1 angle = 180 degrees

Mathematics

2 answers:

4vir4ik [10]3 years ago

3 0

Answer:

It's called a straight angle, it's a straight line.

Step-by-step explanation:

Literally a straight line.

Lunna [17]3 years ago

3 0

Answer:

Straight angle ..

You might be interested in

Determine whether each graph represents a linear or nonlinear function. Use the

ANTONII [103]

Answer:

x-intercepts: -3, -2, -1, 1, 2
y-intercept: 12
symmetry: none

Step-by-step explanation:

The x- and y-intercepts are the points where the graph crosses the x- and y-axes, respectively. If an odd-degree polynomial function has any symmetry, it will be symmetrical about the origin.

<h3>X-intercepts</h3>

The graph shows crossings of the x-axis at -3, -2, -1, 1, and 2. These are the x-intercepts. Expressed as coordinate pairs, each will have a y-coordinate of zero:

(-3, 0), (-2, 0), (-1, 0), (1, 0), (2, 0)

<h3>Y-intercepts</h3>

If the function is not vertically scaled, the y-intercept of an odd-degree function will be the opposite of the product of the x-intercepts:

-(-3)(-2)(-1)(1)(2) = 12

It will have an x-coordinate of zero:

(0, 12) . . . y-intercept

A function can have at most 1 y-intercept.

<h3>Symmetry</h3>

There are two features of this graph that tell you it is an odd-degree function:

the end behavior is in opposite directions
there are an odd number of x-intercepts (x-axis crossings)

An odd-degree function will only be symmetrical about the origin. This function has more negative x-intercepts than positive ones, so is not symmetrical about the origin.

The graph has no symmetry.

<em>Additional comment</em>

Turning points without an x-axis-crossing on either side, or axis "touches" (not crossing) signify complex or even-multiplicity roots. The function will only be of odd degree if there are an odd number of x-axis crossings.

The end behavior of an odd-degree function will be in opposite directions: the infinity will match the sign of x if the leading coefficient is positive, or be opposite to the sign of x if the leading coefficient is negative.

7 0

1 year ago

I need help on this question as well.

Y_Kistochka [10]

8 is the answer, because 4 x 8 = 32

7 0

3 years ago

Read 2 more answers

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)