Find plane Π parallel to 5y-5x-5z=4 containing point p=(2,3,4)
Any plane parallel to 5y-5x-5z=4 has the form 5y-5x-5z=k, or on simplification, y-x-z=k.
Since p lies in the plane, the coordinates of p must satisfy y-x-z=k, or
k=3-2-4=-3
=>
π : y-x-z=-3
equivalently,
π : 5y-5x-5z=-15
Answer:
n=radical of 28
Step-by-step explanation:
n^2=bc
n^2=28
n=radical of 28
Answer:
23) 0
24) Option A
Step-by-step explanation:
For question 23, the line is horizontal. All horizontal lines have no slope.
For question 24, we are only given two points. (-3, -2) and (3, 4).

The slope is 1. Since option A's equation has a slope of one, which is represented by only having the 'x', Option A should be the correct answer.
Easy
1.
Vcyliner=(h)(π)(r²)
2.
Vcone=(1/3)(h)(π)(r²)
3.
Vsphere=(4/3)(π)(r³)
4.
Vcylinder
h=16
r=d/2=8/2=4
Vclinder=(16)(π)(4²)=256π cm³, aprox 804.247 cm³
5.
h=8
r=5
Vcone=(1/3)(8)(π)(5²)=(200π)/3 cm³ aprox 209.43946 cm³
6.
r=10
Vsphere=(4/3)(π)(10³)=(4000π)/3 mi³ aprox 4188.789333 mi³