There+are+no+rational+numbers+that+are+also+whole+numbers

2 answers:

meriva3 years ago

8 0

Actually all the whole numbers are kind of rational numbers like : 1/1 , 2/1 , 3/1 , ...
so there are some rational numbers that are whole numbers too :))
i hope this be helpful
have a nice day

Ilya [14]3 years ago

4 0

No. ALL whole numbers are rational. And what's with all the plus signs ? ? They really make your questions hard to read.

You might be interested in

Trevor measured the town library and made a scale drawing. In real life, the parking lot is 75 meters long. It is 135 centimeter

jasenka [17]

Answer: $9\ cm\equiv 5\ m$

Step-by-step explanation:

Given

The actual length of the parking lot is 75 m

length in the drawing is 135 cm

Here, 75 m is equivalent to 135 cm

Using unitary method

$75\ m\equiv 135\ cm$

$15\ m\equiv 27\ cm$

$9\ cm\equiv \frac{15}{3}=5\ \text{m}$

8 0

2 years ago

Rewrite the following as a single logarithm. Please show ALL work!<br><br> 4logx+logy-1/2logz

fomenos

Answer:
$\log\left(\frac{x^4y}{\sqrt{z}}\right)$
(if this font is too small then try zooming in, or you can check out the attached image which shows the answer in bigger font)

---------------------------------

Work Shown:

I'm going to use the following rules
Rule 1: log(x)+log(y) = log(x*y)
Rule 2: log(x)-log(y) = log(x/y)
Rule 3: y*log(x) = log(x^y)
Rule 4: x^(1/2) = sqrt(x) ... where 'sqrt' stands for 'square root'

Using those rules, we get...
4*log(x)+log(y) - (1/2)*log(z)
log(x^4)+log(y) - log(z^(1/2)) <<-- using rule 3
log(x^4)+log(y) - log(sqrt(z)) <<-- using rule 4
log(x^4*y) - log(sqrt(z)) <<-- using rule 1
log[ (x^4*y)/(sqrt(z)) ] <<-- using rule 2