Can someone please solve both of these showing all the steps I'll add extra points because its two
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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)
</span>
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
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector