You move to the right, so the exponent is gonna be negative.
1.21 * 10-²
Slope is rise over run, so that would be 25/100 = 1/4
Slope is 1/4
There are many ways of solving a system of equation, a very used one is by elimination of variables, which consists in eliminating one of the variables by combining the equations.
Let's do the following combination: (eq 1) + (eq 2)
This means that we add each term vertically:
⇒
When I said "add vertically" means that you add the first term of the first equation with the first term of the second equation. The second term of the first equation with the second term of the second equations, and so on.
This allowed us to remove the "y" temporarily so we can get a linear one-variable equation we can easily solve:
Now that we know the value of "x", we can plug it in any of the starting equations and get "y":
Therefore, our answer looks like this:
9514 1404 393
Answer:
g = 12
Step-by-step explanation:
You recognize that 144 is 12², so you check to see if the x-coefficient is 2·12. It is. The quadratic is a square:
(g -12)² = 0
The only solution is the one that makes the squared term equal to zero:
g -12 = 0
g = 12
_____
The square of a binomial is ...
(a +b)² = a² +2ab +b²
Here, we have a=x, b=-12.
<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>