How does the value of the digit 8 in 1.8 compare to the value of the digit 8 in 486?
1 answer:
Answer:
see the procedure
Step-by-step explanation:
we know that
so
The value of digit 8 in 1.8 is 0.8
The value of digit 8 in 486 is 80
Divide the numbers
80/0.8=100
therefore
The value of the digit 8 in 486 is 100 times greater than the value of digit 8 in 1.8
or
The value of the digit 8 in 1.8 is 100 times less than the value of digit 8 in 486
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Answer:
A)
B)
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are :
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
On top: (-2)³ = -2 × -2 × -2 = -8
(x²)³ = x^6 (the exponents multiply)
and of course, (y)³ = y³
On the bottom: (xy²z)² = x² y^4 z²
(-yz)² = y²z²
Multiplying these together, the exponents add and we get x² y^6 z^4.
So, your reasoning is correct for what you have so far.
Your next step would be cancelling shared factors from the top and bottom.
Just like with regular fractions, if the numerator and denominator are divisible by the same number, you can divide them by it to simplify. (ex: 4/6 = 2/3)
Well, x^6 and x^2 are both divisible by x^2, right?
We can also cancel the y^3.
It might help to visualise the factors like this:
Once you've cancelled out x² and y³ from each, you're left with