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olganol [36]
3 years ago
8

I need to find the arc of GFE, next, I need to find the circumference AND area with a radius of 5 mm. Then the final questions a

sk to Write the equation of a circle with a center at (-1,2) and a diameter of 12.
I will be very thankful for your help, this is a required assignment of mine and I have been struggling to get it done. Thank you :)

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
5 0

Arc GHE is 40 + 80 or 120 so arc GFE is 360 (total measurement in a circle) - 120 which is 240. The circumference of a circle is 2*pi*r so in this case it will be 2*pi*5 or 10pi (you can also write it as approximately 31.4). The area of a circle is pi*r² so it'll be pi*5² or 25pi (you can write it as approximately 78.5 also). The equation of a circle is (x-h)² + (y-k)² = r² where (h,k) is the center of the circle and r is the radius. Input your values. The equation of this circle is (x+1)² + (y-2)² = 6² (The radis is 6 because the diameter is 12)

I hope this helps!

Nina [5.8K]3 years ago
4 0

A full circle is 360 degrees.

You are given the angles for GH, HE and FE, subtract those from 360 to find the angle for FG:

360 - 110 - 80 - 40 = 130 degrees.

Now for the arc GFE add FG and FE:

Arc GFE = 130 + 110 = 240 degrees.

Circumference = 2 x PI x r

Using 3.14 for PI:

Circumference  = 2 x 3.14 x 5 = 31.4 mm or 10PI

Area = PI x r^2 = 3.14 x 25 = 78.5 mm^2 or 25PI mm^2

Equation of a circle with center at (-1,2) and diameter of 12:

The equation is written as (x-x1)^2 + (y-y1)^2 = r^2

x1 and y1 are the values of the center (-1,2) and r is the radius, which would be half the diameter.

The equation is: (x+1)^2 + (y-2)^2 = 36

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Leni [432]

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3 years ago
¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para
Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

\frac{x-2}{p}=\frac{2-y}{p}

y

\frac{x-1}{1-p}=\frac{y-2}{2}

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

\frac{x-2}{p}=\frac{2-y}{p}

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

m_{1}=-1

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

m_{2}=-\frac{1}{m_{1}}

m_{2}=-\frac{1}{-1}

m_{2}=1

ahora tomamos la segunda ecuación y encontramos su pendiente. Tomemos la ecuación:

\frac{x-1}{1-p}=\frac{y-2}{2}

y despejemos y, comenzamos multiplicando ambos lados de la ecuación por 2, así que obtenemos:

2\frac{x-1}{1-p}=y-2

Multiplicamos el 2 por cada término de la fracción, entonces obtenemos:

\frac{2x-2}{1-p}=y-2

ahora pasamos el 2 a sumar al lado izquierdo y obtenemos:

\frac{2x-2}{1-p}+2=y

Ahora podemos separar la fracción del lado izquierdo en dos fracciones para obtener:

\frac{2x}{1-p}-\frac{2}{1-p}+2=y

volteamos la ecuación y nos da:

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de aquí podemos determinar nuestra pendiente:

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Con la primera ecuación determinamos que esta pendiente debería de ser igual a 1, entonces igualamos esa segunda pendiente a 1 para obtener:

\frac{2}{1-p}=1

y despejamos p

Pasamos a multiplicat el 1-p al lado derecho de la ecuación para obtener:

2=1-p

volteamos la ecuación:

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-p=1

y multiplicamos ambos lados de la ecuación por -1 para obtener:

p=-1

Entonces la respuesta es C) solo para el -1

4 0
3 years ago
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