Answer:
Im pretty sure the answer in $8.70
Step-by-step explanation:
All you have to do is 382 divided by 44
Answer:
14
Step-by-step explanation:
19 - 5 = 14
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,

Solution: We are given that a basket contains three green apples and six red apples.
Three apples are randomly selected from the basket.
Now the probability of selecting first apple green is 
The probability of selecting second apple green is 
The probability of selecting third apple green is 
Therefore, the probability of selecting three green apples is:
or 1.19%
Now how many red apples must be added to the basket in order to make the above probability smaller than 0.1%
If we add 11 red apples to the basket, then there will be 17 red apples and 3 green apples in the basket. Therefore, the probability of selecting three green apples if three apples are randomly selected is:
or 0.09%
Therefore, we need to add 11 red apples to the basket to make this probability smaller than 0.1%
Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is −6?
Parallel, slope is the same, perpendicular slope is opposite and reciprocal
so
line n, with slope −6 , same slope -----> Parallel
line p, with slope 1/6 , opposite and reciprocal -----> Perpendicular
line q, with slope −1/6 , -----> Neither
line m, with slope 6, -----> Neither