Answer:8192,-65536,524288
Step-by-step explanation:
2,-16,128,-1024,....
First term=a=2
Common ratio=r=-16/2=-8
Using the formula
Tn=a x r^(n-1)
T5=2 x (-8)^(5-1)
T5=2x (-8)^4
T5=2x4096
T5=8192
T6=2x(-8)^(6-1)
T6=2x(-8)^5
T6=2x-32768
T6=-65536
T7=2x(-8)^(7-1)
T7=2x(-8)^6
T7=2x262144
T7=524288
Answer:
width=25/3
Length=75/3
Step-by-step explanation:
Given
Perimeter=50
Let width=x
Length=4x
So perimeter =2(length+width)
2(length+width)=50
2(X+2x)=50
3x=25
X=25/3
So width=25/3
Length=75/3
Answer:
What is the Pythagorean triplet of 32?
But 32/2 = 16 which is even so:
But 32/4 = 8 which is even so:
But 32/8 = 4 which is even so:
Answer:
Answer in explanation below.
Step-by-step explanation:
32² => 1024 => 1024/4 => 256 => 256 + 1 = 257, 256 - 1 = 255: 32, 255, 257
16² => 256 => 256/4 => 64 => 64 + 1 = 65, 64 - 1 = 63: 2 x ( 16, 63, 65) = 32, 126, 130
8² => 64 => 64/4 => 16 => 16 + 1 = 17, 16 - 1 = 15: 4 x ( 8, 15, 17) = 32, 60, 68
4² => 16 => 16/4 => 4 => 4 + 1 = 5, 4 - 1 = 3: 8 x ( 4, 3, 5) = 32, 24, 40
What is the exact circumference of the circle?
Answer: c=2πr
Step-by-step explanation:
because im just that guy
Answer:
c=62.8
Step-by-step explanation:
diameter=20
radius=10
c=2πr
c=2×22/7×10
c=62.8
Answer:
0.25
Step-by-step explanation:
Let
r = red
b = blue
g = green
y = yellow
From the 4 + 4 + 4 + 4 = 16 possible outcomes, there are 4 in which the chips are of the same color (rr, bb, gg, yy). The probability that the two chips selected are the same color is
4/16 = 0.25 = 25%
Salt flows into the tank at a rate of
(5 lb/gal) * (6 gal/min) = 30 lb/min
The volume of solution in the tank after t min is
600 gal + (6 gal/min - 12 gal/min)*(t min) = 600 - 6t gal
which means salt flows out at a rate of
(A(t)/(600 - 6t) lb/gal) * (12 gal/min) = 2 A(t)/(100 - t) lb/min
Then the net rate of change of the salt content is modeled by the linear differential equation,
Solve for A:
Multiply both sides by the integrating factor, :
Integrate both sides:
The tank starts with no salt, so A(0) = 0 lb. This means
and the particular solution to the ODE is