Answer:
we have (a,b,c)=(4,-2,0) and R=4 (radius)
Step-by-step explanation:
since
x²+y²+z²−8x+4y=−4
we have to complete the squares to finish with a equation of the form
(x-a)²+(y-b)²+(z-c)²=R²
that is the equation of a sphere of radius R and centre in (a,b,c)
thus
x²+y²+z²−8x+4y=−4
x²+y²+z²−8x+4y +4 = 0
x²+y²+z²−8x+4y +4 +16-16 =0
(x²−8x + 16) + (y² + 4y + 4 ) + (z²) -16 = 0
(x-4)² + (y+2)² + z² = 16
(x-4)² + (y-(-2))² + (z-0)² = 4²
thus we have a=4 , b= -2 , c= 0 and R=4
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
Answer:
The value of x is: x > -2
Step-by-step explanation:
The interval notation if you need it is (-2, ♾)
Answer:
㏒e(x)= 5
Step-by-step explanation:
