Answer:
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,

b is write answers click it
Answer:
26 ft.2
Step-by-step explanation
Divide 104 by 4 because the sandbox is a square. With a square all sides are equal length, so you just divide by the amount of sides.
Hope this helps. :)
First, to make things easier, rewrite the equation
3(65+7)+2
2. distribute the 3 inside the parentheses
3 * 65 = 195 3 * 7 = 21
3. add the products together
195 + 21 = 216
4. add 216 to the remaining term, 2.
216 + 2 = 218.
the answer is 218
Wendy spent four hours on the train.
If the...
Train Speed= 60 mph
Train Time= x
Bus Speed= 40 mph
Bus Time= 5.5-x
Total Time= 5.5 hr
and
Total Distance= 300 miles
Then we can put all of that into the equation...
60x + 40(5.5-x)= 300
Now, following the order of operations we can free the x from the parenthesis by multiplying 40 by 5.5 and (-x) to get...
60x + (-40x) + 220= 300
Next we can simplify by adding the 60x and (-40x) to get...
20x + 220= 300
After that we need to get the x alone on the left side. To do this we must use reverse order; if we subtract 220 from 220 on the left side and then we can subtract 220 from 300 on the left side. Remember the Golden Rule of Algebra! If you add, subtract, multiply, divide, or raise to a power on one side you must do it on the other! This gives us...
20x + 220 - 220= 300 - 220
20x= 80
Now all that is left to get x all alone on the left is to look back at the golden rule of algebra and divide 20 from both 20x and 80. Which gives us...
20x/20 = 80/20
x= 4
And that gives us our answer! 4 hours :D
Hope this helped!