mprjug6
23.02.2020 •
Mathematics
Expression is X plus X plus X plus X plus X represents the number of beers and four bags which expression is equivalent to the expression above be sure to show your work
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Ответ:
X + X + X + X + X = N + 4B,
5X = N + 4B.
Here, N denotes the number of beers and B denotes bag.
Step-by-step Explanation:
Expression X plus X plus X plus X plus X is mathematically written as:
X + X + X + X + X = 5X
Let's denote the number of beers as N
Also, let's denotes bag as B.F our Bags can then be written as 4B
Therefore, The expression X plus X plus X plus X plus X represents the number of beers and four bags is equivalent to:
5X = N + 4B
Ответ:
∫
e
x
sin
2
x
d
x
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
+
C
Explanation:
We seek the integral:
I
=
∫
e
x
sin
2
x
d
x
There is no suitable substitution, however, We can apply Integration By Parts:
Let
{
u
=
sin
2
x
⇒
d
u
d
x
=
2
cos
2
x
d
v
d
x
=
e
x
⇒
v
=
e
x
Then plugging into the IBP formula:
∫
(
u
)
(
d
v
d
x
)
d
x
=
(
u
)
(
v
)
−
∫
(
v
)
(
d
u
d
x
)
d
x
We have:
∫
(
sin
2
x
)
(
e
x
)
d
x
=
(
sin
2
x
)
(
e
x
)
−
∫
(
e
x
)
(
2
cos
2
x
)
d
x
∴
I
=
e
2
x
sin
2
x
−
2
∫
e
x
cos
2
x
d
x
Now consider the integral given by:
I
2
=
∫
e
x
cos
2
x
d
x
We will now need to apply IBP again:
Let
{
u
=
cos
2
x
⇒
d
u
d
x
=
−
2
sin
2
x
d
v
d
x
=
e
x
⇒
v
=
e
x
Then plugging into the IBP formula we have::
∫
(
cos
2
x
)
(
e
x
)
d
x
=
(
cos
2
x
)
(
e
x
)
−
∫
(
e
x
)
(
−
2
sin
2
x
)
d
t
I
2
=
e
x
cos
2
x
+
2
∫
e
x
sin
2
x
d
t
=
e
x
cos
2
x
+
2
I
And so combining the results we find that:
I
=
e
x
sin
2
x
−
2
{
e
x
cos
2
x
+
2
I
}
=
e
x
sin
2
x
−
2
e
x
cos
2
x
−
4
I
∴
5
I
=
e
x
(
sin
2
x
−
2
cos
2
x
)
∴
I
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
And not forgetting the constant of integration,
I
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
+
C∫
e
x
sin
2
x
d
x
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
+
C
Explanation:
We seek the integral:
I
=
∫
e
x
sin
2
x
d
x
There is no suitable substitution, however, We can apply Integration By Parts:
Let
{
u
=
sin
2
x
⇒
d
u
d
x
=
2
cos
2
x
d
v
d
x
=
e
x
⇒
v
=
e
x
Then plugging into the IBP formula:
∫
(
u
)
(
d
v
d
x
)
d
x
=
(
u
)
(
v
)
−
∫
(
v
)
(
d
u
d
x
)
d
x
We have:
∫
(
sin
2
x
)
(
e
x
)
d
x
=
(
sin
2
x
)
(
e
x
)
−
∫
(
e
x
)
(
2
cos
2
x
)
d
x
∴
I
=
e
2
x
sin
2
x
−
2
∫
e
x
cos
2
x
d
x
Now consider the integral given by:
I
2
=
∫
e
x
cos
2
x
d
x
We will now need to apply IBP again:
Let
{
u
=
cos
2
x
⇒
d
u
d
x
=
−
2
sin
2
x
d
v
d
x
=
e
x
⇒
v
=
e
x
Then plugging into the IBP formula we have::
∫
(
cos
2
x
)
(
e
x
)
d
x
=
(
cos
2
x
)
(
e
x
)
−
∫
(
e
x
)
(
−
2
sin
2
x
)
d
t
I
2
=
e
x
cos
2
x
+
2
∫
e
x
sin
2
x
d
t
=
e
x
cos
2
x
+
2
I
And so combining the results we find that:
I
=
e
x
sin
2
x
−
2
{
e
x
cos
2
x
+
2
I
}
=
e
x
sin
2
x
−
2
e
x
cos
2
x
−
4
I
∴
5
I
=
e
x
(
sin
2
x
−
2
cos
2
x
)
∴
I
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
And not forgetting the constant of integration,
I
=
1
5
e
x
(
sin
2
x
−
2
cos
2
x
)
+
C