1. Exercise 18, section 5.4. Compute 90, 91, 92, 93, 94 and 95. Make a conjecture about the units digit of 9n where n is a posit
1 answer:
Answer:
n is a positive integer and here 90, 91, 92, 93, 94 and 95 is 9⁰ = 1
9¹ = 9
9² = 81
9³ = 729
9⁴ = 6561
9⁵ = 59049
so mathematical induction is prove given below.
Step-by-step explanation:
9⁰ = 1
9¹ = 9
9² = 81
9³ = 729
9⁴ = 6561
9⁵ = 59049
lets us consider by P (n) for the base case n = 0 then 9⁰ = 1 So P (1) is true
k ≥ 1so consider all integers
1≤l≤ k
There is a need to prove P (k+1)
if k is odd
=
.9.
is 9 so
= 9 +10m
then
= 9 ( 9 + 10m)
=81 + 10 (9m)
so digit is 1 and k +1 is even
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Answer:
9a²b
Step-by-step explanation:
Hi there!
We need to find the greatest common factor out of 45a²b³ and 18b
We can split apart the monomials to make it easier
45a²b³ is 45*a²b³
18b is 18*
b
First, let's find the GCF out of 45 and 18 (the number coefficients)
we can find all of the multiples of the 2 numbers:
45 is made up of 9 and 5
9 is made up of 3 and 3
so 3*3*5 is 45
18 is made up of 2 and 9
9 is made up of 3 and 3
so 2*3*3 is 18
3*3 is in both 45 and 18, so 9 is the GCF out of 45 and 18
Now let's find the GCF out of a²b³ and b
a²b³ made up of a² and b³
so a²b³ is a*a*b*b*b
b is made up of
and b
so b is a*a*a*a*b
a*a*b is in both a²b³ and b, so the GCF out of a²b³ and
b is a²b
Now multiply 9 and a²b together, as they are only the GCF of the parts of the monomials
9*a²b=<u>9a²b</u>
there's the greatest common factor of the 2 monomials
Hope this helps!