Each colection day: D
Number of tops collected on that day: N
D1=1; N1=2
D2=3; N2=8
1) Linear model
N-N1=m(D-D1)
m=(N2-N1)/(D2-D1)
m=(8-2)/(3-1)
m=(6)/(2)
m=3
N-N1=m(D-D1)
N-2=3(D-1)
N-2=3D-3
N-2+2=3D-3+2
N=3D-1
when D=6:
N=3(6)-1
N=18-1
N=17
<span>What is the number of tops collected on the sixth day based on the linear model?
</span>The number of tops collected on the sixth day based on the linear model is 17.
2) Exponential model
N=a(b)^D
D=D1=1→N=N1=2→2=a(b)^1→2=ab→ab=2 (1)
D=D2=3→N=N2=8→8=a(b)^3→8=a(b)^(1+2)
8=a(b)^1(b)^2→8=ab(b)^2 (2)
Replacing (1) in (2)
(2) 8=2(b)^2
Solving for b:
8/2=2(b)^2/2
4=(b)^2
sqrt(4)=sqrt( b)^2 )
2=b
b=2
Replacing b=2 in (1)
(1) ab=2
a(2)=2
Solving for a:
a(2)/2=2/2
a=1
Then, the exponential model is N=1(2)^D
N=(2)^D
When D=6:
N=(2)^6
N=64
<span>What is the number of tops collected on the sixth day based on the exponential model?
</span><span>The number of tops collected on the sixth day based on the exponential model is 64</span>
Answer:
(x/2)-16=f( x)
Step-by-step explanation:
f( x)=2x+32
y=2x+32
inve.rse (switc.h x an.d y)
x=2y+32
subtr.act 32 on eac.h sid.e
x-32=2y
divid.e ea.ch s.ide b.y 2
(x-32)/2=(2y)/2
(x/2)-16=y
rev.ise y a.s f( x)
(x/2)-16=f( x)
So.me wo.rd is sup.posedly ba.d an.d I do.n't kn.ow wha.t s.o I pu.t ra.ndom per.iods eve.rywhere.
Answer:
5.5 units
Step-by-step explanation:
Arc length is given by ...
s = rθ
s = (3.5)(π/2) ≈ 5.5 . . . units
Answer:
-1 is not in the domain of (f o g)(x)
Step-by-step explanation:
f(x) = sqrt(x - 9)
g(x) = -6x - 3
(f o g)(x) = f(g(x)) = sqrt(g(x) - 9)
(f o g)(x) = sqrt(-6x - 3 - 9)
(f o g)(x) = sqrt(-6x - 12)
Let x = -1:
(f o g)(-1) = sqrt(-6(-1) - 12)
(f o g)(-1) = sqrt(6 - 12)
(f o g)(-1) = sqrt(-6)
Since sqrt(-6) is not a real number, -1 is not in the domain of (f o g)(x).
