Algebra 1A unit 6 lesson 6
1 answer:
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The standard form of a circle is:
Where
(h, k) is the center
r is the radius
Since the center is (1, 1), we can write:
EG (hypotenuse of the right triangle) is the radius of the circle. Thus, we can say >>>
The correct answer is
BAnswer:
The area can be written as
And the value of it is approximately 1.8117
Step-by-step explanation:
x = u/v
y = uv
Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.
- x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
- x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4
Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).
Lets compute the partial derivates of x and y over u and v
Therefore, the Jacobian matrix is
and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))
In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:
Therefore,