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never [62]
3 years ago
6

Geometry help please?

Mathematics
2 answers:
valentinak56 [21]3 years ago
7 0
A) All internal angles of a square add up to 360° , just like a rectangle.
b) All sides are equal in length.
c) They both have 2 pairs of parallel sides.
d) Each side is equal in length.
scZoUnD [109]3 years ago
4 0
A) All angles are right angles, opposite sides are parallel, opposite sides are congruent
b) All sides are congruent
c) opposite sides are parallel
d) all sides are congruent
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D) 28 is the right answer
find the attachments for the details

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It translates to addition.
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The ratio of the sides of a triangle is 2:6:7.
tangare [24]

suppose the three sides are 2x, 6x, and 7x in length.

2x+6x+7x+195

15x=195

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7 0
2 years ago
Does anyone know how to do this? I’m confused
nikklg [1K]

Answer:

cos(θ)

Step-by-step explanation:

Para una función f(x), la derivada es el límite de  

h

f(x+h)−f(x)

​

, ya que h va a 0, si ese límite existe.

dθ

d

​

(sin(θ))=(  

h→0

lim

​

 

h

sin(θ+h)−sin(θ)

​

)

Usa la fórmula de suma para el seno.

h→0

lim

​

 

h

sin(h+θ)−sin(θ)

​

 

Simplifica sin(θ).

h→0

lim

​

 

h

sin(θ)(cos(h)−1)+cos(θ)sin(h)

​

 

Reescribe el límite.

(  

h→0

lim

​

sin(θ))(  

h→0

lim

​

 

h

cos(h)−1

​

)+(  

h→0

lim

​

cos(θ))(  

h→0

lim

​

 

h

sin(h)

​

)

Usa el hecho de que θ es una constante al calcular límites, ya que h va a 0.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)(  

h→0

lim

​

 

h

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)

Para calcular el límite lim  

h→0

​

 

h

cos(h)−1

​

, primero multiplique el numerador y denominador por cos(h)+1.

(  

h→0

lim

​

 

h

cos(h)−1

​

)=(  

h→0

lim

​

 

h(cos(h)+1)

(cos(h)−1)(cos(h)+1)

​

)

Multiplica cos(h)+1 por cos(h)−1.

h→0

lim

​

 

h(cos(h)+1)

(cos(h))  

2

−1

​

 

Usa la identidad pitagórica.

h→0

lim

​

−  

h(cos(h)+1)

(sin(h))  

2

 

​

 

Reescribe el límite.

(  

h→0

lim

​

−  

h

sin(h)

​

)(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

−(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

Usa el hecho de que  

cos(h)+1

sin(h)

​

 es un valor continuo en 0.

(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)=0

Sustituye el valor 0 en la expresión sin(θ)(lim  

h→0

​

 

h

cos(h)−1

​

)+cos(θ).

cos(θ)

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