PLEASE HELP ASAP!!! IM GONNA FAIL
1 answer:
Answer:
Explained.
Step-by-step explanation:
The graph of a line on the coordinate system represents the points that are on the graph that will satisfy the equation of the line.
Now, if two lines on the coordinate plane are graphed and they pass through the same point (h,k) that means the point satisfies both the equations of the lines.
Let us have two curve equations and
and they pass through the same point (h,k) on the coordinate plane.
Then we can write ......... (1) and
.......... (2)
Now, solving equations (1) and (2) we get
........... (3)
Therefore, we have to solve the above equation (3) to get the value of h i.e. the x-coordinate of the point/s where the graph of the equations (1) and (2) intersect.
Now, converting h to x we will get the same result. (Answer)
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<span>For the plane, we have z = 5x + 9y
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
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For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals