(z+1.2)/5=3.6
quotient: division
Answer:
A. a parabola that opens down.
Step-by-step explanation:
First we can eliminate two answers, b and d, since parabalas that open left or right use the y variable.
Now the equation of a standard parabola is x². This means that there are no negative y values, and therefore it cannot open down, it is up.
With a NEGATIVE coefficient, the graph is flipped over the x axis, therefore making it open down.
Graphical proof is attached for further understanding, positive a is in red, negative a is in blue.
Answer:
- The car's rate in kilometers per hour is 108 kilometers per hour
- The car will travel 540 kilometers in 5 hours
Step-by-step explanation:
Here is the complete question:
A car is traveling at a rate of 30 meters per second. What is the car's rate in kilometers per hour? How many kilometers will the car travel in 5 hours? Do not round your answers.
Step-by-step explanation:
- To determine the car's rate in kilometers per hour, we will convert 30 meters per second to kilometer per hour.
1000 meters = 1 kilometer
∴ 1 meter = 0.001 kilometer
3600 seconds = 1 hour
∴ 1 second = 1/3600 hour
Therefore,
= 
= 
= 108 kilometers / hour
Hence, the car's rate in kilometers per hour is 108 kilometers per hour.
- To determine how many kilometers the car will travel in 5 hours, that is we are to determine the distance covered by the car in 5 hours.
From the formula
Speed = Distance / Time
∴ Distance = Speed × Time
(NOTE: Speed is also known as rate)
Speed (Rate) = 108 kilometers / hour
Time = 5 hours
∴ Distance = Speed × Time gives
Distance = 108 × 5
Distance = 540 kilometers
Hence, the car will travel 540 kilometers in 5 hours.
Answer:
3 to 1
Step-by-step explanation:
Because the 6 is the bigger number but is also on the left which means the number on the right has to divide into the six therefore 3 to 1
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)