A blue container weighs 4 x 103
1 answer:
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The given inequality is:
This inequality can be divided in two parts as:
a)
b)
Solving part a:
Solving part b:
Therefore, the solution to the given inequality is
and 
. Combining both the ranges we get the solution: 
.
In interval notation, this solution can be expressed as [1,5]
This inequality can be divided in two parts as:
a)
b)
Solving part a:
Solving part b:
Therefore, the solution to the given inequality is
In interval notation, this solution can be expressed as [1,5]
We first obtain the equation of the lines bounding R.
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by:
![63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576](https://tex.z-dn.net/?f=63%20%5Cint%5Climits%5E1_0%20%5Cint%5Climits%5E%7B1%7D_0%20%7B%285u-23v%29%7D%20%5C%2C%20dudv%20%3D63%20%5Cint%5Climits%5E1_0%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Du%5E2-23uv%5Cright%5D%5E%7B1%7D_0%20%5C%5C%20%20%5C%5C%20%3D63%5Cint%5Climits%5E1_0%28%5Cfrac%7B5%7D%7B2%7D-23v%29dv%3D63%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Dv-%5Cfrac%7B23%7D%7B2%7Dv%5E2%5Cright%5D%5E1_0%3D63%5Cleft%28%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B23%7D%7B2%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D63%28-9%29%3D%7C-576%7C%3D576)
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by: