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kodGreya [7K]
4 years ago
14

Solve for x, m/1, and m/2.​

Mathematics
1 answer:
ki77a [65]4 years ago
4 0
The answers would be M/3
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Apply the distributive property to create an equivalent expression. ( 1 − 2 g + 4 h ) ⋅ 5 =
mina [271]

Answer:

{(1-2g) + 4h] x 5 =

Step-by-step explanation:

8 0
3 years ago
Which expression is equivalent to 54.5-6?​
pentagon [3]

Answer: 5-6•54

Hope this helps (:

3 0
3 years ago
Abdullah expands (x - 3)2 and gets the answer x2 - 9. Describe what he has<br> done wrong.
solniwko [45]

Answer:

He squared the numbers inside without following the appropriate rule

Step-by-step explanation:

<u>what he should have done </u>

<u />(x-3)^2\\= x^2 - 2 *x*3+3^2\\= x^2-6x+9

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3 years ago
Las llantas de un automóvil pequeño tiene una medida de 14 pulgadas de diámetro.Realiza la gráfica de una de las llantas colocan
LiRa [457]

Answer:

Invitamos cordial a revisar la imagen adjunta abajo para mayor detalle sobre la gráfica de las llantas.

La forma general de la circunferencia de la llanta está representada por x^{2}+y^{2}-49 = 0.

Step-by-step explanation:

A continuación, anexamos una representación de las llantas del automóvil como una circunferencia centrada en el origen y con un diámetro de 14 pulgadas, es decir, un radio de 7 pulgadas.

La ecuación estándar de la circunferencia centrada en un punto dado y con un radio determinado está definida por:

(x-h)^{2}+(y-k)^{2} =r^{2} (1)

Donde:

h, k - Coordenadas del centro de la circunferencia, medidas en pulgadas.

r - Radio de la circunferencia, medido en pulgadas.

Si conocemos que h = 0\,in, k = 0\,in y r = 7\,in, entonces la ecuación estándar que representa a la circunferencia de la llanta es:

x^{2}+y^{2} = 49

Ahora, la ecuación general de la circunferencia centrada en el origen satisface la siguiente expresión:

A\cdot x^{2}+A\cdot y^{2}+B = 0 (2)

Donde A, B son los coeficientes de forma de la circunferencia.

Entonces, la forma general de la circunferencia de la llanta está representada por la siguiente expresión:

x^{2}+y^{2}-49 = 0

6 0
3 years ago
As she drove down the icy Road Miss Campbell slammed on her breaks her car did a 360 explain what happened to her car ​
meriva

Answer:

The friction that the car usually uses to stop(a rough and hard ground) was null and void, so the car spun uncontrollably.

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3 years ago
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