What is the midpoint of the coordinate F( -2, 4) and G(6, 1)

Mathematics

1 answer:

Marrrta [24]2 years ago

6 0

Answer:

$\huge\boxed{\left(2; 2\dfrac{1}{2}\right)}$

Step-by-step explanation:

$\text{The formula of a midpoint:}\\\\M\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{We have the points}\ F(-2,\ 4),\ G(6,\ 1).\\\\\text{Substitute:}\\\\M_{FG}\left(\dfrac{-2+6}{2};\ \dfrac{4+1}{2}\right)\to M_{FG}\left(\dfrac{4}{2};\ \dfrac{5}{2}\right)\to M_{FG}\left(2; 2\dfrac{1}{2}\right)$

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PLS HELP! WILL GIVE BRAINLIEST

Leviafan [203]

Answer:

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Step-by-step explanation:

The standard form of the equation of a circle is:

( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius

Rewrite the given equation in the standard form:

2x^2 + 2y^2 - 8x + 10y + 2 = 0
x^2 - 4x + y^2 + 5y = -1
x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
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(2, 2.5)

And radius is:

$\sqrt{37}$ / 2

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2 years ago

Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.

alina1380 [7]

Answer:

The answer is $x=a^{-1}cb^{-1}$ .

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$ , where $e$ stands for the identity element of G.