........................................d
<span>Take the integral:
integral (cos(x))/sqrt(cos(x)+1) dx
For the integrand (cos(x))/sqrt(1+cos(x)), substitute u = 1+cos(x) and du = -sin(x) dx:
= integral (u-1)/(sqrt(2-u) u) du
For the integrand (-1+u)/(sqrt(2-u) u), substitute s = sqrt(2-u) and ds = -1/(2 sqrt(2-u)) du:
= integral -(2 (1-s^2))/(2-s^2) ds
Factor out constants:
= -2 integral (1-s^2)/(2-s^2) ds
For the integrand (1-s^2)/(2-s^2), cancel common terms in the numerator and denominator:
= -2 integral (s^2-1)/(s^2-2) ds
For the integrand (-1+s^2)/(-2+s^2), do long division:
= -2 integral (1/(s^2-2)+1) ds
Integrate the sum term by term:
= -2 integral 1/(s^2-2) ds-2 integral 1 ds
Factor -2 from the denominator:
= -2 integral -1/(2 (1-s^2/2)) ds-2 integral 1 ds
Factor out constants:
= integral 1/(1-s^2/2) ds-2 integral 1 ds
For the integrand 1/(1-s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds:
= sqrt(2) integral 1/(1-p^2) dp-2 integral 1 ds
The integral of 1/(1-p^2) is tanh^(-1)(p):
= sqrt(2) tanh^(-1)(p)-2 integral 1 ds
The integral of 1 is s:
= sqrt(2) tanh^(-1)(p)-2 s+constant
Substitute back for p = s/sqrt(2):
= sqrt(2) tanh^(-1)(s/sqrt(2))-2 s+constant
Substitute back for s = sqrt(2-u):
= sqrt(2) tanh^(-1)(sqrt(1-u/2))-2 sqrt(2-u)+constant
Substitute back for u = 1+cos(x):
= sqrt(2) tanh^(-1)(sqrt(sin^2(x/2)))-2 sqrt(1-cos(x))+constant
Factor the answer a different way:
= sqrt(1-cos(x)) (csc(x/2) tanh^(-1)(sin(x/2))-2)+constant
Which is equivalent for restricted x values to:
Answer: |
| = (2 cos(x/2) (2 sin(x/2)+log(cos(x/4)-sin(x/4))-log(sin(x/4)+cos(x/4))))/sqrt(cos(x)+1)+constant</span>
Answer:
<em>Numbers: 6 and -2</em>
Step-by-step explanation:
<u>Equations</u>
This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.
However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:
![x+y=4\qquad\qquad [1]](https://tex.z-dn.net/?f=x%2By%3D4%5Cqquad%5Cqquad%20%5B1%5D)
![x.y=-12\qquad\qquad [2]](https://tex.z-dn.net/?f=x.y%3D-12%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for y:

Substituting in [2]

Operating:

Rearranging:

Solving with the quadratic formula:

With a=1, b=-4, c=-12:



The solutions are:


This confirms the preliminary results.
Numbers: 6 and -2
Answer:
D b = -2E +212
Step-by-step explanation:
The slope is given by -2 for every 1000 ft
The initial value is 212 degrees
We can use
y = mx+b where m is the slope and b is the initial value
y = -2x +212
We are at an elevation of E thousand ft
Replace x with E
y = -2 *E +212
b is the boiling point at this elevation
We replace y with b
b = -2E +212
Answer:
B
Step-by-step explanation:
For a line to be parallel it must have the same slope
6x-2y=-5
+5 +5
6x-2y+5 = 0
+2y +2y
6x+5 =2y
/2 /2
3x+5/2=y So the slope is 3 and we know it can only be either B or D
Now we need to find which line passes through (7,-2) and we find this by plugging 7 into B and D
B. -2= 21-23 TRUE
D. -2 = 21+ 23 FALSE
So the answer is B