Answer:
See attachment for graph
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Step-by-step explanation:
Given
![f(x) = \left[\begin{array}{cc}-1&x1\end{array}\right](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26x%3C-1%5C%5C0%26-1%5Cle%20x%20%5Cle%20-1%5C%5C1%26x%3E1%5Cend%7Barray%7D%5Cright)
Required
The graph of the step function
Before plotting the graph, it should be noted that:
and
use closed circle at its end
and
use open circle at its end
So, we have:
![f(x) = -1,\ \ \ \ x < -1](https://tex.z-dn.net/?f=f%28x%29%20%3D%20-1%2C%5C%20%5C%20%5C%20%5C%20x%20%3C%20-1)
The line stops at -1 with an open circle
![f(x) = 0,\ \ \ \ -1 \le x \le 1](https://tex.z-dn.net/?f=f%28x%29%20%3D%200%2C%5C%20%5C%20%5C%20%5C%20-1%20%5Cle%20x%20%5Cle%201)
The line starts at - 1 and stops at -1 with a closed circle at both ends
![f(x) = 1,\ \ \ \ x > 1](https://tex.z-dn.net/?f=f%28x%29%20%3D%201%2C%5C%20%5C%20%5C%20%5C%20x%20%3E%201)
The line starts at 1 with an open circle
The options are not complete, so I will plot the graph myself.
<em>See attachment for graph</em>
Given:
The expression is
![\dfrac{8}{3}-\dfrac{5}{6}-\dfrac{2}{12}](https://tex.z-dn.net/?f=%5Cdfrac%7B8%7D%7B3%7D-%5Cdfrac%7B5%7D%7B6%7D-%5Cdfrac%7B2%7D%7B12%7D)
To find:
The expression which is equivalent to the given expression.
Solution:
We have,
![\dfrac{8}{3}-\dfrac{5}{6}-\dfrac{2}{12}](https://tex.z-dn.net/?f=%5Cdfrac%7B8%7D%7B3%7D-%5Cdfrac%7B5%7D%7B6%7D-%5Cdfrac%7B2%7D%7B12%7D)
It can be written as
![=\dfrac{8}{3}-\dfrac{5}{6}-\dfrac{1}{6}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B8%7D%7B3%7D-%5Cdfrac%7B5%7D%7B6%7D-%5Cdfrac%7B1%7D%7B6%7D)
Taking LCM, we get
![=\dfrac{16-5-1}{6}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B16-5-1%7D%7B6%7D)
![=\dfrac{10}{6}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B10%7D%7B6%7D)
![=\dfrac{5}{3}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B5%7D%7B3%7D)
The mixed fraction of this improper fraction is
![=\dfrac{1\times 3+2}{3}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1%5Ctimes%203%2B2%7D%7B3%7D)
![=1\dfrac{2}{3}](https://tex.z-dn.net/?f=%3D1%5Cdfrac%7B2%7D%7B3%7D)
So, the expression
is equivalent to the given expression.
Therefore, the correct option is A.
It would be the second one because since it’s at 5. You shift from 5 to 8 three units to the right
Answer:
We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489
Step-by-step explanation:
- Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.
This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.
However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.
Observe that:
- 1+2+4+8+16+32+64+128+256+489=1000
- Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
- If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since
, by 2) we know that this would be possible.
Answer:
Always Irrational
Step-by-step explanation:
An Irrational number is described as a number which cannot in the actual sense be expressed as a ratio between two integers and is not an imaginary number. It means that an irrational number cannot be expressed as a simple fraction.
If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without repetition. Hence irrational numbers are floating point numbers.
The quotient of a rational number and an irrational number is always irrational.