Find the intercepts for both planes.
Plane 1, <em>x</em> + <em>y</em> + 2<em>z</em> = 2:



Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:



Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.
Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.
So the volume of the region is



Answer:
equation of the circle J with center K(-7,3) and radius 6 is=
x^2+y^2+14x-6y+22=0
Step by step explanation:
Equation of circle with given centre (h,k) and radius r is
(x-h)^2+(y-k)^2=r^2
Given centre are K(-7,3) and radius 6.
So equation of Circle=
(x- -7)^2+(y-3)^2=6^2
x^2+14x+49+y^2-6y+9=36
x^2+y^2+14x-6y+22=0 answer
Answer:
B
Step-by-step explanation:
The measure of the arc is equal to the central abgle it subtends, thus
arc = 65°
Right angles = 90 degrees
8c - 5 + 6c - 3 = 90
14c - 8 = 90
14c = 98, c = 7
Solution: c = 7