Answer:
Point E on the number line represents the approximate volume of a cylinder with a radius of 4 units and height of 4 units.
Step-by-step explanation:
It is given that the radius of the cylinder is 4 units and the height of the cylinder is 4 units.
The volume of the cylinder is
Where, r is radius and h is height of the cylinder.
Substitute r=4, h=4, π=3.14 in the above formula.
The volume of the cylinder is 200.96. Therefore, point E on the number line represents the approximate volume of a cylinder with a radius of 4 units and height of 4 units.
Answer:
Time should be plotted on the horizontal X - axis.
Step-by-step explanation:
Xavier and Bobby each receive a salary check every week. They combine their checks in one account. Their account has $100 after 2 weeks and $300 after 6 weeks.
In a graph showing the relationship of amount in their account versus time, time should be plotted on the horizontal X - axis.
Answer/Step-by-step explanation:
well, first figure out how many units/times prism A was enlarged to become prism B...
16 ÷ 2 = 8
Now if we apply that info to the length of the prism, then...
80 ÷ 8 = 10
Therefore x = 10
And the Prisms remain congruent and proportional!
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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)
Put 7 in place of x 10-2(7)= 10-14=negative 4
