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How do you order 6.037, 6.37, 6.5, 6.3772 from least to greatest?

EZ as pi
Featured 5 months ago

$6.037 , \text{ "6.37," "6.3772," } 6.5$

Explanation:

Write the number with the same number of decimal places first.
The most number of decimal places is $4$, so use that for all of them.

This makes comparing them much easier. Look at the units digits first: They are all $6$, so at this point they are all the same.

$\textcolor{b l u e}{6} .0370$
$\textcolor{b l u e}{6} .3700$
$\textcolor{b l u e}{6} .5000$
$\textcolor{b l u e}{6} .3772$

Look at the first decimal place next (tenths)

$\textcolor{b l u e}{6} \textcolor{red}{.0} 370 \text{ } \leftarrow$ this one is the smallest $\left(0\right)$
$\textcolor{b l u e}{6} .3700$
$\textcolor{b l u e}{6} \textcolor{red}{.5} 000 \text{ } \leftarrow$ this one is the biggest $\left(5\right)$
$\textcolor{b l u e}{6} .3772$

For the two remaining numbers - look at the second decimal place:
They are both $\textcolor{m a \ge n t a}{7}$, so the two numbers are equal at this point

$\textcolor{b l u e}{6} .3 \textcolor{m a \ge n t a}{7} 00$
$\textcolor{b l u e}{6} .3 \textcolor{m a \ge n t a}{7} 72$

Now look at the third decimal place:

$\textcolor{b l u e}{6} .3 \textcolor{m a \ge n t a}{7} \textcolor{\lim e g r e e n}{0} 0 \text{ } \leftarrow$ this is smaller $\left(0\right)$
$\textcolor{b l u e}{6} .3 \textcolor{m a \ge n t a}{7} \textcolor{\lim e g r e e n}{7} 2 \text{ } \leftarrow$ this is bigger $\left(7\right)$

Now we can order the numbers from smallest to biggest:

$\textcolor{b l u e}{6} \textcolor{red}{.0} 370 , \text{ "color(blue)(6).3color(magenta)(7)color(limegreen)(0)0," "color(blue)(6).3color(magenta)(7)color(limegreen)(7)2," } \textcolor{b l u e}{6} \textcolor{red}{.5} 000$

Now you can write them as the original values.

$6.037 , \text{ "6.37," "6.3772," } 6.5$

Find the GCF?

Serena D.
Featured 4 months ago

$2 \left(4 x + 1\right)$

Explanation:

The GCF of 8x and 2 is 2.

To find the GCF, first find the prime factorization of each term.

The prime factorization for 8x is 2, 2, 2, and x (those 4 terms multiply to equal 8x).

The prime factorization for 2 is 2 and 1.

To find the GCF of these two terms, multiply the common factor(s). In this case, since the only common factor is 2, that is the GCF.

Now divide each term by 2 and put the resulting expression within parentheses and put the 2 on the outside.

$2 \left(4 x + 1\right)$

What fraction is equivalent to 1/3?

EZ as pi
Featured 4 months ago

$\frac{1}{3} = \frac{2}{6} = \frac{3}{9} = \frac{4}{12} = \frac{5}{15} = \frac{9}{27} = \frac{14}{42} = \frac{17}{51} = \frac{50}{150} \ldots . .$

Explanation:

There are many fractions which are equivalent to $\frac{1}{3}$ but this is the simplest form.

To find an equivalent fraction, multiply $\frac{1}{3}$ by $1$, but with $1$ written as $\frac{2}{2} , \frac{3}{3} , \frac{4}{4}$ etc

Multiplying the numerator and denominator by the same number does not change the value of a fraction.

$\frac{1}{3} \times \frac{2}{2} = \frac{2}{6}$

$\frac{1}{3} \times \frac{7}{7} = \frac{7}{21}$

$\frac{1}{3} = \frac{2}{6} = \frac{3}{9} = \frac{4}{12} = \frac{5}{15} = \frac{9}{27} = \frac{14}{42} = \frac{17}{51} = \frac{50}{150}$ etc

How you test 69,902 for divisibility by 2, 3, 5, 9, or 10?

Shwetank Mauria
Featured 3 months ago

$69902$ is divisible only by $2$ and not by $3 , 5 , 9$ or $10$. Please see below for details.

Explanation:

Divisibility by $2$ can be tested by checking the digit at unit's place. If we have $0 , 2 , 4 , 6$ or $8$ at unit's place, then the number is divisible by $2$. Here we have $2$ at unit's place in $69902$, and hence it is divisible by $2$.

Divisibility by $3$ can be tested by checking that sum of all the digits is divisible by $3$ or not. Here sum of digit's in $69902$ is $6 + 9 + 9 + 0 + 2 = 26$, which is not divisible by $3$, therefore $69902$ is not divisible by $3$.

Divisibility by $5$ can be tested by checking the digit at unit's place. If we have $0$ or $5$ at unit's place, then the number is divisible by $5$. Here we have $2$ at unit's place in $69902$, and hence it is not divisible by $5$.

Divisibility by $9$ can be tested by checking that sum of all the digits is divisible by $9$ or not. Here sum of digit's in $69902$ is $6 + 9 + 9 + 0 + 2 = 26$, which is not divisible by $9$, therefore $69902$ is not divisible by $9$. Note that if a number is not divisible by $3$, it is also not divisible by $9$.

Divisibility by $10$ can be tested by checking the digit at unit's place. If we have $0$ at unit's place, then the number is divisible by $10$. Here we have $2$ at unit's place in $69902$, and hence it is not divisible by $10$.

How do you find the prime factorization of 196?

Shwetank Mauria
Featured 3 months ago

$196 = 2 \times 2 \times 7 \times 7$

Explanation:

As the last two digits in $196$ are divisible by $4$, $196$ too is divisible by $4$.

Dividing by $4$ we get $49$ and hence

$196 = 4 \times 49$

but factors of $4$ are $2 \times 2$ and that of $49$ are $7 \times 7$. Further $2 ' s$ and $7 ' s$ are prime numbers and cannot be factorized.

Hence prime factors of $196$ are

$196 = 2 \times 2 \times 7 \times 7$

Note : This method of factorization, in which we first find identifiable factors and then proceed until all prime factors are known is called tree method. This is graphically described below.

What is the least common multiple of 7 and 24?

Tony B
Featured 2 months ago

A teacher will expect the prime number method. Just for the hell of it this is a different approach!

168

Explanation:

We have two numbers ; 24 and 7

I am going to count the 24's. However lets look at this value.

24 can be 'split' into a sum of 7's with a remainder. So each 24 consists of:

$24 = \left(7 + 7 + 7 + 3\right)$

If we sum columns of these we will get the 3 summing to a value into which 7 will divide exactly. When this happens we have found our least common multiple.

REMEMBER WE ARE COUNTING THE 24's

$\text{ count "color(white)("dd") "The 24's}$
$\textcolor{w h i t e}{\text{ddd") 1color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 2color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 3color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 4color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 5color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 6color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
#color(white)("ddd") 7color(white)("ddd")ul( (color(white)(.)7+color(white)(.)7+color(white)(.)7+color(white)(.)3)larr" Add"#
$\textcolor{w h i t e}{\text{ddddddd.}} 49 + 49 + 49 + \underbrace{21}$
$\textcolor{w h i t e}{\text{dddddddddddddddddddd}} \downarrow$
#color(white)("dddddddddddddd")" exactly divisible by 7"#

We have a count of 7 so the value is $7 \times 24 = 168$

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