Is the centroid of a triangle equidistant from the three vertices?

Mathematics

1 answer:

Xelga [282]2 years ago

4 0

The point that is equidistant to all sides of a triangle is called the incenter.

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Find the x-intercepts of the parabola with vertex (1,-108) and y-intercept (0,-105). Write your answer in this form: (X1,y1), (x

Stolb23 [73]

Answer:

(7, 0) and (-5, 0)

Step-by-step explanation:

Vertex form

$y=a(x-h)^2+k$

(where (h, k) is the vertex)

Given:

vertex = (1, -108)

$\implies y=a(x-1)^2-108$

Given:

y-intercept = (0, -105)

$\implies a(0-1)^2-108=-105$

$\implies a(-1)^2=-105+108$

$\implies a=3$

Therefore:

$\implies y=3(x-1)^2-108$

The x-intercepts are when y = 0

$\implies 3(x-1)^2-108=0$

$\implies 3(x-1)^2=108$

$\implies (x-1)^2=36$

$\implies x-1=\pm \sqrt{36}$

$\implies x=1\pm 6$

$\implies x=7, x=-5$

Therefore, the x-intercepts are (7, 0) and (-5, 0)

7 0

2 years ago

Graph the first six terms of a sequence where a1 = −10 and d = 3.

Neko [114]

notice the first term is -10, and the common difference is 3, so we add 3 to get the next term.

$\bf \begin{array}{cccll}\stackrel{x}{n}&term&\stackrel{y}{value}\\\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\1&a_1&-10\\\\2&a_2&\stackrel{-10+3}{-7}\\\\3&a_3&\stackrel{-7+3}{-4}\\\\4&a_4&\stackrel{-4+3}{-1}\\\\5&a_5&\stackrel{-1+3}{2}\\\\6&a_6&\stackrel{2+3}{5}\end{array}$