Substitute x = 3 - 2 cos(θ) and dx = 2 sin(θ) dθ (where "sin" = "sen"). So we have
∫ sin(θ) / (3 - 2 cos(θ)) dθ = 1/2 ∫ 1/x dx
= 1/2 ln|x| + C
= 1/2 ln(3 - 2 cos(θ)) + C
(We can remove the absolute value because -1 ≤ cos(θ) ≤ 1, so 1 ≤ 3 - 2 cos(θ) ≤ 5, and |x| = x when x ≥ 0.)
number.
The sum of a rational number and an irrational number is
an (A:a rational) or (B: a irrational)
number.
~~~ZoomZoom44~~~~
Answer:
Its B
Step-by-step explanation:
Answer:
The angle between them is 60 degrees
Step-by-step explanation:
Given
Required
The angle between them
The cosine of the angle between them is:
First, calculate a.b
Multiply the coefficients of like terms
Next, calculate |a| and |b|
Recall that:
This gives:
Take arccos of both sides
Answer:
to find the right answer multiple by 2
Step-by-step explanation:
and you get 14/16
What percent of the carbon-14 had been lost from the pollen?
(half-life of carbon-14 = 5730)
Answer:
inf(A) does not exist.
Step-by-step explanation:
As per the question:
We need to prove that A is closed under multiplication,
If for every
Proof:
Suppose, x, y
Since, both x and y are real numbers thus xy is also a real number.
Now, consider another set B such that:
B = {xy} has only a single element 'xy' and thus [B] is bounded.
Since, [A] represents the union of all the bounded sets, therefore,
⇒ xy
Therefore, from x, y , we have xy .
Hence, set a is closed under multiplication.
Now, to prove whether inf(A) exist or not
Proof:
Let us assume that inf(A) exist and inf(A) =
Thus is also a real number.
Let C be another set such that
C = { - 1}
Now, we know that C is a bounded set thus { - 1} is also an element of A
Also, we know:
inf(A) =
Therefore,
But
is an element of A and
This is contradictory, thus inf(A) does not exist.
Hence, proved.