Therefore, a(s) = 1 + (−3)2 + (−2)2 da d = 14 da. d since we want the area of the section of the plane that lies above the first octant, then the region d will be the triangular region bounded by the x-axis, the y-axis, and the line formed by the intersection of the plane with the plane z = 0. substituting z = 0 into 3x + 2y + z = 6 and solving for y, we find that this line of intersection has the equation

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

180-142=38 for the acute angle next to the entry

90+38=128

180-128=52

a=52

step-by-step explanation: