V=4/3πr^3
36π=4/3πr^3
3/4(36π)=(4/3πr^3)3/4
27π=πr^3
π(27π)=r^3
27=r^3
square root
r=3
s.a=4πr^2
s.a=4π(3^2)
s.a=4π(9)
s.a=36π
Answer:
C, 5/2
Step-by-step explanation:
3 I2x -1I + 4 = 16
Isolate absolute value:
3 I2x -1I + 4 - 4 = 16 - 4
3 I2x -1I= 12
(3 I2x -1I) / 3 = 12 / 3
I2x -1I= 4
Since the question is only asking for the positive, all you need to solve for is the +4. Solve for x
2x -1 = 4
2x -1 + 1 = 4 +1
2x = 5
2x / 2 = 5 / 2
x = 5/2 or C
This kind of triangle is acute triangle because on the given three angles, none of them does not exceed more than 90°.
So there is a real and imaginary axis
the midopint is just the average of them
average between the reals is (5-3)/2=2/2=1
average between imaginaries is (18i+2i)/2=20i/2=10i
center is 1+10i
The required plane Π contains the line
L: (-1,1,2)+t(7,6,2)
means that Π is perpendicular to the direction vector of the line L, namely
vl=<7,6,2>
It is also required that Π be perpendicular to the plane
Π 1 : 5y-7z+8=0
means that Π is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.
Thus the normal vector of the required plane, Π can be obtained by the cross product of vl and vp, or vl x vp:
i j k
7 6 2
0 5 -7
=<-42-10, 0+49, 35-0>
=<-52, 49, 35>
which is the normal vector of Π
Since Π has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
Π : -52(x-(-1))+49(y-1)+35(z-2)=0
=>
Π : -52x+49y+35z = 171
Check that normal vector of plane is orthogonal to line direction vector
<-52,49,35>.<7,6,2>
=-364+294+70
=0 ok
