It is not factored completely because it was only factored by 3ab but it could have been factored by 6a²b
6a²b(a+6+8a²)
Answer:
1)
Area
2)
1/2
3)
Radius
Step-by-step explanation:
We know that for any circle with radius r the formula for the area and circumference of the circle is given as:
Area of circle=πr².
and circumference of circle = 2πr.
Hence, to describe the relationship between the area of a circle and its circumference.
The Area of circle is 1/2 times the radius times the circumference.
( Since,
Area=1/2 × r× 2πr
Area=πr² )
Answer:
6 whole packages as the other half can't be count
Step-by-step explanation:
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)
