The IQR is the range of the box in the box and whisker plot
So if you remember how to do the average of something you are already there
You need 2 numbers
First being the lower quartile and upper quartile
48 is where the lower quartile starts and 70 is where the upper quartile finishes *NEVER USE MEDIAN IN IQR
all you have to do is subtract
70-40 and you get as an answer 22.0 being A as your answer
27j-6=14j+7
Subtract 14j from both sides
27j-6-14j=14j+7-14j
13j-6=7
Add 6 to both sides
13j-6+6=7+6
13j=13
Divide both sides by 13
13j/13=13/13
j=1
I hope that's help !
Let's see.
We have function 
. We have to determine if the function is even or odd or neither.
We can find this out in many ways but the assignment specifically asks to prove this algebraically.
Function is even if 
, so, 
Which is true since negative raised to an even power is positive to that same number.
Similarly, function is odd if 
, that is,
.
So we have prooved that function 
is an even function not an odd function and therefore also not neither.
Hope this helps.
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)
