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DerKrebs [107]
3 years ago
12

What is the equation, in point-slope form, for a line that goes through (8, −4) and has a slope of −56 ?

Mathematics
2 answers:
olga55 [171]3 years ago
6 0

The point-slope form:

y-y_1=m(x-x_1)

We have the point (8, -4) and the slope m = -56. Substitute:

y-(-4)=-56(x-8)\\\\\boxed{y+4=-56(x-8)}

nikklg [1K]3 years ago
6 0

y + 4 = -5/6 (x + 8)

(i had the same question on a k12 quiz)

if its actually -56 then the equation would be  

y + 4 = -56 (x+8)

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Answer:

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Step-by-step explanation:

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4 0
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1. Una escalera de 4 m de longitud se apoya sobre una pared vertical. Si la distancia entre la base de la escalera a la pared es
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Answer:

puedes resolver de dos maneras si diste teorema de Pitágoras lo aplicas

hipotenusa al cuadrado = cateto 2 +cateto 2 ( el 2 significa al cuadrado)

sustituyes

hipotenusa= 4m

cateto= 2,5

hay que hallar el otro cateto que nos daría la altura a la que está la escalera

despejamos y tenemos

cateto 2= hip2 -cat2

cat2=(4)2-(2,5)2=

       =   16-6,25=9,75

luego hallamos la raíz cuadrada de 9,75= 3,12

     

la altura a la que se encuentra es 3,12m

-si aun no diste Pitágoras podes representarlo en una hoja utilizando cm en lugar de metros( a escala). trazas el triángulo rectángulo la base te la da la distancia a la cual se encuentra la escalera de la pared es decir 2,5cm trazas la hipotenusa de 4cm de manera que coincida con el cateto opuesto , y mides el valor de este ,será de 3,12cm no olvides que la respuesta la debes dar en metros ya que es la unidad de medida que te da.

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Change the expression to a single square root, or its opposite:
aleksley [76]

Answer:

a)2\sqrt{2}=\sqrt{8}

b)-7\sqrt{3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b}  =\sqrt{2.b}

d)5\sqrt{y} =\sqrt{25y}

e)-6\sqrt{2a}  =-\sqrt{72a}

f)-0.1\sqrt{200c}=-  \sqrt{2c

Step-by-step explanation:

a) 2\sqrt{2}=( \sqrt{2}  )^2.\sqrt{2}=(\sqrt{2})^3 = \sqrt{2^3} =\sqrt{8}

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e)-6\sqrt{2a} =-\sqrt{6^2}\sqrt{2a}  = -\sqrt{36.2a} =-\sqrt{72a}

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4 0
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