Alicia drew a pentagon

Mathematics

2 answers:

Verdich [7]3 years ago

4 0

Good for her what do you want me to answer

babunello [35]3 years ago

4 0

A pentagon has 5 sides, and 5 central angles, and all the sides and all the angles are the same.

What else did you want to know?
Each central angle is 360/5 degrees, or 72 degrees.

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On a coordinate plane, a line is drawn from point K to point J. Point K is at (160, 120) and point J is at (negative 40, 80). Wh

emmainna [20.7K]

The x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is (85, 105).

Given

On a coordinate plane, a line is drawn from point K to point J. Point K is at (160, 120) and point J is at (negative 40, 80).

<h3>Coordinates</h3>

The coordinates point any point can be found by using the following formula.

$\rm X=\dfrac{m}{m+n}(x_2-x_1)+x_1\\\\ Y=\dfrac{m}{m+n}(y_2-y_1)+y_1$

The x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is;

$\rm X=\dfrac{m}{m+n}(x_2-x_1)+x_1\\\\ X = \dfrac{3}{3+5}(-40-160)+160\\\\ X = 3 \times (-25)+160\\\\\X = -75+160\\\\ X = 85\\\\\ Y=\dfrac{m}{m+n}(y_2-y_1)+y_1\\\\ Y=\dfrac{3}{3+5}(80-120)+120\\\\Y =- 3\times 5 +`120\\\\Y=-15+120\\\\Y=105$

Hence, the x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is (85, 105).

To know more about co-ordinates click the link given below.

brainly.com/question/13847533

5 0

2 years ago

h(x)= \dfrac{1}{8} x^3 - x^2h(x)= 8 1 x 3 −x 2 h, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided b

Nesterboy [21]

Answer:

$\dfrac{1}{2}$

Step-by-step explanation:

Given the function $h(x)=\dfrac{1}{8}x^3-x^2$

The average rate of change of the function h(x) over the interval $a\le x\le b$ can be calculated using formula

$\dfrac{h(b)-h(a)}{b-a}$

In your case,

$a=-2,\\ \\b=2,$

so the average rate of change is

$\dfrac{h(2)-h(-2)}{2-(-2)}\\ \\=\dfrac{(\frac{1}{8}\cdot 2^3-2^2)-(\frac{1}{8}\cdot (-2)^3-(-2)^2)}{4}\\ \\=\dfrac{(1-4)-(-1-4)}{4}\\ \\=\dfrac{-3+5}{4}\\ \\=\dfrac{2}{4}\\ \\=\dfrac{1}{2}$