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VMariaS [17]
2 years ago
12

Move point B some more. As you move point B, the angle formed between A ⁢ B ↔ and C ⁢ D ↔ varies. If you want to make A ⁢ B ↔ pe

rpendicular to C ⁢ D ↔ what do you need to do? Explain in terms of ∠ ⁢ B ⁢ E ⁢ C .
Mathematics
1 answer:
Scrat [10]2 years ago
6 0

The Explanation in terms of  ∠ ⁢ B ⁢ E ⁢ C  is that to make AB perpendicular to CD, one has to move point B till  the angle between the lines, BEC is said to measures 90 degree.

<h3>What are  perpendicular angles?</h3>

A perpendicular angle is known to be an angle that is seen on a straight line and it is said to be one that tends to makes an angle of 90° with any other line.

Note that 90° is also known to be a right angle and it is one that is often denoted or marked by a little square that tend to exist between two perpendicular lines.

Therefore, the two lines are said to intersect at a right angle, and so are known to be perpendicular to one another.

Hence, The Explanation in terms of  ∠ ⁢ B ⁢ E ⁢ C  is that to make AB perpendicular to CD, one has to move point B till  the angle between the lines, BEC is said to measures 90 degree.

Learn more about perpendicular angles from

brainly.com/question/1202004

#SPJ1

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Resolver las siguientes inecuaciones cuadráticas y presentar el conjunto solución en forma de intervalo y gráficamente. 1) X2 &l
Blizzard [7]

Respuesta:

1) (-5,2)

2) (-\infty,-2]U(\frac{1}{2},\infty)

Explicación paso a paso:

1)

x^{2}

Para comenzar este problema, debemos moverlo todo al lado izquierdo de la inecuación, por lo que obtenemos:

x^{2}+3x-10

Ahora podemos factorizar el lado izquierdo para obtener:

(x+5)(x-2)

Ahora podemos cambiar el símbolo < por un = para encontrar los valores de x en los cuales la inecuación es igual a cero.

(x+5)(x-2)=0

Y luego despejamos x.

x+5=0

x=-5

y

x-2=0

x=2

Ahora construimos nuestros intervalos posibles.

(-\infty,-5)

(-5,2)

y

(2,\infty)

Y escogemos algunos valores de prueba. Estos nos ayudarán a determinar si cada intervalo hace que la inecuación sea verdadera o falsa.

(-\infty,-5)

Para este intervalo escojamos -6 y evaluemoslo en la inecuación.

(x+5)(x-2)<0

(-6+5)(-6-2)<0

(-1)(-8)<0

8<0

falso, así que este intervalo no es parte de nuestra respuesta.

(-5,2)

para este escojamos x=0 y probémoslo en la inecuación.

(x+5)(x-2)<0

(0+5)(0-2)<0

(5)(-2)<0

-10<0

verdadero, así que este intervalo es parte de nuestra respuesta.

(2,\infty)

Para este escojamos 3 y probémoslo en unestra inecuación.

(x+5)(x-2)<0

(3+5)(3-2)<0

(8)(1)<0

8<0

falso, así que este intervalo no es parte de nuestra respuesta.

así que nuestra respuesta es:  (-5,2)

Vea imagen adjunta para representación gráfica.

2)

2x^{2}+3x\geq2

Para resolver este problema, comenzamos moviéndolo todo al lado izquierdo de la inecuación.

2x^{2}+3x-2\geq0

Ahora podemos factorizar el lado izquierdo de la inecuación para obtener:

(2x-1)(x+2)\geq0

Ahora podemos cambién el símbolo ≥ por un símbolop de = para obtener los valores de x que hacen que la inecuación sea igual a 0.

(2x-1)(x+2)=0

y ahora despejamos x.

2x-1=0

x=\frac{1}{2}

y

x+2=0

x=-2

Ahora construimos nuestros intervalos posible.

(-\infty,-2]

[-2,\frac{1}{2}]

y

[\frac{1}{2},\infty)

Ahora escogemos los valores de prueba correspondientes.

(-\infty,-2]

para este, escojamos -3 y probémoslo en la inecuación.

(2x-1)(x+2)\geq0

(2(-3)-1)(-3+2)\geq0

(-7)(-1)\geq0

7\geq0

verdadero, así que este intervalo es parte de nuestra respuesta.

[-2,\frac{1}{2}]

para este, utilicemos 0 como valor de prueba.

(2x-1)(x+2)\geq0

(2(0)-1)(0+2)\geq0

(-1)(2)\geq0

-2\geq0

falso, así que este intervalo no es parte de nuestra respuesta.

[\frac{1}{2},\infty)

para este, utilicemos 1 como valor de prueba.

(2x-1)(x+2)\geq0

(2(1)-1)(1+2)\geq0

(1)(3)\geq0

3\geq0

verdadero, así que este intervalo es parte de nuestra respuesta.

Así que nuestra respuesta es la unión entre los dos intervalos que resultaron verdadero, por lo que nuestra respuesta es:

(-\infty,-2]U[\frac{1}{2},\infty)

Vea la representación gráfica en la imagen adjunta.

6 0
3 years ago
Which expressions are equivalent to 4(4a+5)?
SCORPION-xisa [38]

Answer:

B

Step-by-step explanation:

Distribute the 4

4(4a+5) = 4·4a + 4·5 = 16a + 20

6 0
3 years ago
Consider the algebraic expression:
GrogVix [38]
Coefficients are the numbers that are next to a variable(a letter in an expression). Therefore the coefficients would be 7 and 18 because they are both next to a variable (y and x).

I can’t exactly remember what a constant is but I think it is a number alone (not next to any variables) and they are ONLY separated by + and - not multiplication or division. In this case it would be all of the numbers that are in between the + or - (all of them.)

I also think a constant could be the variable next to the coefficient. Which would be y and x (next to 7 and 18).

I’m am so sorry I can’t remember what a constant is but I hope the coefficient helps:)
7 0
4 years ago
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Need Help Plzz!!!!!!!!!!!!!!
antoniya [11.8K]

Answer:

EFD

Step-by-step explanation:

AC/DE = AB/EF = BC/DF

So the correct order should be EFD

6 0
3 years ago
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What are the factors of x^2-144?
Ksju [112]

Answer:

(x-12) (x+12)

Step-by-step explanation:

If you factor it out since both are perfect squares you would place an x on each side and then find the square root of 144, which is 12 and put a positive and a negative in each parenthesis.

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