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24wilsleaann
09.05.2021 •
Mathematics
Write an explicit rule for the nth term in the sequence 2,6,18,54
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Ответ:
1.h(x)
2.i(x)
3.m(x)
4.k(x)
5.g(x)
6.l(x)
7.j(x)
8.f(x)
Step-by-step explanation:
Given:
l(x)= lim x→∞ 5x^2-4/x^2+1
Dividing by highest power of x i.e. x^2
l(x)= lim x→∞ (5-4/x^2)/(1+1/x^2)
As per infinity property of 4/x^2 and 1/x^2 = 0
l(x)= 5/1
l(x)=5
i(x)= lim x→∞ x-1/|1-4x|
Dividing by highest power of x i.e. x
i(x)= lim x→∞ (1-1/x) / |1/x-4|
As per infinity property of 1/x and 1/x = 0
i(x)= -1/4
i(x)=-o.25
f(x)=lim x→∞ x^2-1000/x-5
Dividing by highest power of x i.e. x^2
f(x)= lim x→∞ (1-1000/x^2) / (1/x-5/x^2)
As per infinity property of 1000/x^2 and 1/x and 5/x^2 = 0
f(x)= 1/0
f(x)=∞
m(x)=lim x→∞ 4x^2-6/1-4x^2
Dividing by highest power of x i.e. x^2
m(x)= lim x→∞ (4-6/x^2) / (1/x^2-4)
As per infinity property of 6/x^2 and 1/x^2 = 0
m(x)= -4/4
m(x)= -1
g(x)=lim x→∞ |4x-1|/x-4
Dividing by highest power of x i.e. x
g(x)= lim x→∞ (|4-1/x|) / (1-4/x)
As per infinity property of 1/x and 4/x = 0
g(x)= 4/1
g(x)= 4
h(x)=lim x→∞ x^3-x^2+4/1-3x^3
Dividing by highest power of x i.e. x^3
h(x)= lim x→∞ (1-1/x+4/x^3)/(1/x^3-3)
As per infinity property of 1/x ,4/x^3 and 1/x^3 = 0
h(x)= -1/3
h(x)= -0.333
k(x)=lim x→∞ 5x+1000/x^2
Dividing by highest power of x i.e. x^2
k(x)= lim x→∞ (5/x+1000/x^2)/(1)
As per infinity property of 5/x ,1000/x^2 = 0
k(x)= 0/1
k(x)=0
j(x)=lim x→∞ x^2-1/|7x-1|
Dividing by highest power of x i.e. x^2
j(x)= lim x→∞ (1-1/x^2)/(|7/x-1/x^2|)
As per infinity property of 1/x^2 ,1/x^2 and 7/x = 0
j(x)= 1/0
j(x)=∞
functions in ascending order will be:
h(x)
i(x)
m(x)
k(x)
g(x)
l(x)
j(x)
f(x) !