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2 years ago

Is it possible for a number to be a rational number that is not an interger but is a whole number. Explain

Mathematics

1 answer:

mylen [45]2 years ago

5 0

Whole numbers are a subset of integers, which in turn are a subset of rational numbers.

So, every whole number is an integer, and every integer is a rational number.

So, it is possible for a rational number not to be an integer. Think of any decimal number: 1.356 is a rational number, but it's not an integer.

On the other hand, if a number is not an integer, it can't be a whole number, because all whole numbers are integers.

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how many 1/3 inch cubes does it take to fill a box with width 2 2/3 inches length 3 1/3 inches and height 2 1/3 inches

Alona [7]

Answer:

560/9 or around 63 cubes

Step-by-step explanation:

2 2/3 -> 8/3

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2 1/3-> 7/3

7/3 x 8/3= 56/9

56/9 x 10/3 = 560/27

560/27 divided by 1/3

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AlekseyPX

Answer:

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Step-by-step explanation:

There are two clear points shown on the graph

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1 year ago

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Kisachek [45]

Answer:

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Step-by-step explanation:

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2 years ago

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Consider the square root function f(x) = sqrt x + 2 + 1 to complete the table of values below. find each missing value and then

Scrat [10]

Given the function :

$f(x)=\sqrt[]{x+2}+1$

We need to find each missing value

Given x = -3 , -2 , -1 , 2 , 7

So, substitute with each value of x to find the corresponding value of f(x)

$x=-3\rightarrow f(x)=\sqrt[]{-3+2}+1=\sqrt[]{-1}+1$

So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)

$\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\ \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\ \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\ \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}$

the graph of the function and the points will be as shown in the following image :

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4 months ago