Answer:
The rational numbers are 
and the irrational functions are 
.
Step-by-step explanation:
A rational number can be expressed in the form of 
, where p and q are integers and q is not equal to 0. For example 
.
An irrational function can not be expressed in the form of , where p and q are integers and q is not equal to 0. For example 
.
If any number is multiplied by a irrational number then the resultant number is an irrational number.
By the above definition we can conclude that:
The number 
is a rational number.
Therefore 
is an irrational number.
Therefore 6.25 is a rational number.
Therefore 0.01045 is a rational number.
The number 
is a rational number.
The number 
is an irrational number.
Therefore 
is an irrational number. The numbers with recursive bar are always rational numbers.
Answer:
seven and sixty-four thousandths
Step-by-step explanation:
in factors it would be
7 x 1
+ 0 x 0.1
+ 6 x 0.01
+ 4 x 0.001
Answer:
g = 21/4
Decimal form:
g = 5.25
Mixed Number:
5 1/4
1. The number of sample size 1 jelly beans in a 2-liter jar is <u>645</u>.
2. The number of sample size 2 jelly beans in a 2-liter jar is <u>640</u>.
3. The number of sample size 3 jelly beans in a 2-liter jar is <u>637</u>.
<h3>What is a mathematical operation?</h3>
A mathematical operation is an expression involving the use of mathematical operands and operators to compute values.
Mathematical operations use variables, numbers, and operators (addition, subtraction, division, and multiplication).
<h3>Data and Calculations:</h3>
Total weight = 1,150g
Weight of the jar = 440g
The total weight of the jelly beans = 710g (1,150 - 440)
Sample Size 1: the number of jelly beans = 645 (710/22.0 x 20)
Sample Size 2: the number of jelly beans = 640 (710/22.2 x 20)
Sample Size 3: the number of jelly beans = 637 (710/22.3 x 20)
Thus, the number of jelly beans in a 2-liter jar depends on the sample size of the jelly beans.
Learn more about mathematical operations at brainly.com/question/20628271
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Answer:
:)
Step-by-step explanation:
There are several ways to do this. The first, which I personally think is the quickest way without having to do too much math or rearranging the equation is to simply pick random points for x and solve for y.
For instance, you might try -2 for x.
- ( - 2)^2 - 4 ( - 2) - 3
= - 4 + 8 - 3
= 1
So one of your points would be (-2, 1)
Then you might choose, say, -3 for x. And after that, -1 for x. If you want to graph even more points after those (the more points, the more accurate your graph), choose -4 and 0 for x. Simply put it into the equation and find y. Then graph.
