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4 years ago

Examples of how to add mixed numbers

2 answers:

6 0

You first change the denominater to common and then you add the whole numbers and then the fractions 1 1/2 + 1 1/4 =1 2/4 +1 1/4+= 2 3/4

suter [353]4 years ago

3 0

1. 2 3/4 + 1 1/4

This problem is a simple addition problem, all you have to do is add 2+1 (3) and then if the denominator of the fractions are the same, then you add freely without changing the denominator. So we add 3/4+1/4 (4/4) If the numerator is equivalent or greater than the denominator; we "Convert". So, since the numerator is equal to the denominator; we will Convert that fraction into a whole number (1) because a fraction is nothing but a part. If all the parts of the fraction are put together, then we will get a whole. After that we add the 1 to the 3 (4). The answer to this problem is 3.

2. 2 3/4 + 2 3/4

This problem is relatively easy. Do the steps above, but you will see that the fraction has turned into a improper fraction (6/4) Now, if you've already subtracted 4 from the numerator, then you've done a great job! Now we add 1 to the whole numbers, making it 5, but we still have work to do. There are still 2 parts left to the fraction, which means we have to make a mixed number. All you do is take the 5 and put 2 over 4, we can reduce the fraction, so let's go ahead and do so. The answer to this problem would be 5 1/2.

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Find the numbers 3% of a number is 18, 3% 1% 100% So the number is

leva [86]

Answer:

Step-by-step explanation:

3% OF a number IS 18

turn ur percent to a decimal

" of " means multiply and " is " means equals

let x represent a number

0.03x = 18

x = 18 / 0.03

x = 600 <===

7 0

3 years ago

Dean is saving up change so he can go to the arcade with 100 quarters. If he has 363 pennies, 200 nickels, and 23 dimes. How muc

solong [7]

D. $9.07
explanation: 363 pennies is equal to $3.36, 200 nickels is equal to $10.00, and 23 dimes is equal to $2.30. all of this change added up is equal to $15.93. since 100 quarters are worth $25, you would take 25-15.93. this equals D.$9.07

8 0

3 years ago

Read 2 more answers

Find an integer x such that 0<=x<527 and x^37===3 mod 527

Greeley [361]

Since $527=17\times31$ , we have that

$x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}$

By Fermat's little theorem, and the fact that $37=2(17)+3=1(31)+6$ , we know that

$x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}$
$x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}$

so we have

$\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}$

Consider the first case. By Fermat's little theorem, we know that

$x^{17}\equiv x^{16}x\equiv x\mod{17}$

so if we were to raise $x^5$ to the $n$ th power such that

$(x^5)^n\equiv x^{5n}\equiv x\mod{17}$

we would need to choose $n$ such that $5n\equiv1\mod{16}$ (because $16+1\equiv1\mod{16}$ ). We can find such an $n$ by applying the Euclidean algorithm:

$16=3(5)+1$
$\implies1=16-3(5)$
$\implies16-3(5)\equiv-3(5)\equiv1\mod{16}$

which makes $-3\equiv13\mod{16}$ the inverse of 5 modulo 16, and so $n=13$ .

Now,

$x^5\equiv3\mod{17}$
$\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}$

$3^1\equiv3\mod{17}$
$3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}$
$3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}$
$\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}$

Similarly, we can look for $m$ such that $7m\equiv1\mod{30}$ . Apply the Euclidean algorithm:

$30=4(7)+2$
$7=3(2)+1$
$\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)$
$\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}$

so that $m=13$ is also the inverse of 7 modulo 30.

And similarly,

$x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}$

Decomposing the power of 3 in a similar fashion, we have

$3^{13}\equiv(3^3)^4\times3\mod{31}$

$3\equiv3\mod{31}$
$3^3\equiv27\equiv-4\mod{31}$
$\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}$

So we have two linear congruences,

$\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}$

and because $\mathrm{gcd}\,(17,31)=1$ , we can use the Chinese remainder theorem to solve for $x$ .

Suppose $x=31+17$ . Then modulo 17, we have

$x\equiv31\equiv14\mod{17}$

but we want to obtain $x\equiv12\mod{17}$ . So let's assume $x=31y+17$ , so that modulo 17 this reduces to

$x\equiv31y+17\equiv14y\equiv1\mod{17}$

Using the Euclidean algorithm:

$17=1(14)+3$
$14=4(3)+2$
$3=1(2)+1$
$\implies1=3-2=5(3)-14=5(17)-6(14)$
$\implies-6(14)\equiv11(14)\equiv1\mod{17}$

we find that $y=11$ is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

$x\equiv31(11)(12)+17\equiv12\mod{17}$

To satisfy the second condition that $x\equiv24\mod{31}$ , taking $x$ modulo 31 gives

$x\equiv31(11)(12)+17\equiv17\mod{31}$

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find $z$ such that $17z\equiv1\mod{31}$ . Euclidean algorithm:

$31=1(17)+14$
$17=1(14)+3$

and so on - we've already done this. So $z=11$ is the inverse of 17 modulo 31. Now, we take

$x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}$

as required. This means the congruence $x^{37}\equiv3\mod{527}$ is satisfied by

$x=31(11)(12)+17(11)(24)=8580$

We want $0\le x$ , so just subtract as many multples of 527 from 8580 until this occurs.

$8580=16(527)+148\implies x=148$

3 0

4 years ago

An interior angle of a regular polygon measures 170 degrees. How many sides does it have?

Leya [2.2K]

$\bf \textit{sum of all interior angles in a polygon}\\\\ n\theta =180(n-2)~~ \begin{cases} n= \stackrel{number~of}{sides}\\ \theta =\stackrel{angle~in}{degrees}\\[-0.5em] \hrulefill\\ \theta =170 \end{cases}\implies n170=180(n-2) \\\\\\ 170n=180n-360\implies 170n+360=180n\implies 360=10n \\\\\\ \cfrac{360}{10}=n\implies 36=n$

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3 years ago

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