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3 years ago

Which number sentence below is not correct? 24.154 < 24.15 23.07 < 23.072 24.67 = 24.6700 28.045 > 28.044

Mathematics

2 answers:

Andrew [12]3 years ago

6 0

24.154 <24.15 that is the incorrect one

crimeas [40]3 years ago

3 0

Answer:

Option 1 - 24.154 < 24.15

Step-by-step explanation:

To find : Which number sentence below is not correct?

Solution :

1) 24.154 < 24.15

It is false as 24.154 is 0.004 times greater than 24.15.

2) 23.07 < 23.072

It is true as 23.07 is less than 23.072.

3) 24.67 = 24.6700

It is true as zero after decimal doesn't make any change.

4) 28.045 > 28.044

It is true as 28.045 is greater than 28.044.

Therefore, Option 1 is correct.

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Which number sentence is not true?

melomori [17]

Answer:

Step-by-step explanation:

The absolute value of a number is the actual distance of the number from zero. So, it is always a positive number. No negative value.

A) I -4.5I = 4.5 TRUE

B) I 0I < I -45I TRUE

Reason: 0 < 45

C) I 45 I > 0 TRUE

D) I4.5 I > I - 45 I FALSE

Reason: 4.5is not greater than 45

5 0

3 years ago

(a) the number 561 factors as 3 · 11 · 17. first use fermat's little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11),

Vitek1552 [10]

LFT says that for any prime modulus $p$ and any integer $n$ , we have

$n^p\equiv n\pmod p$

From this we immediately know that

$a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}$

Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case $11\times17=187=62\times3+1$ , so

$a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3$

Next, $63=21\times3$ , so

$a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3$

Next, $21=7\times3$ , so

$a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3$

Finally, $7=2\times3+1$ , so

$a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3$

We do the same thing for the remaining two cases:

$3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}$

$3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}$

Now recall the Chinese remainder theorem, which says if $x\equiv a\pmod n$ and $x\equiv b\pmod m$ , with $m,n$ relatively prime, then $x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}$ , where ${m_n}^{-1}$ denotes $m^{-1}\pmod n$ .

For this problem, the CRT is saying that, since $a^{561}\equiv a\pmod3$ and $a^{561}\equiv a\pmod{11}$ , it follows that

$a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}$
$\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}$
$\implies a^{561}\equiv 34a\equiv a\pmod{33}$

And since $a^{561}\equiv a\pmod{17}$ , we also have

$a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}$
$\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}$
$\implies a^{561}\equiv562a\equiv a\pmod{561}$

6 0

4 years ago

Please Help Quick <br> 25 Points!

den301095 [7]

1.) D
2.) Option 2
3.) B
4.) B

Make me brainiest pls

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3 years ago

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saw5 [17]

Answer:

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Step-by-step explanation:

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2 years ago

You deposit $750 into an account that earns 8% interest, compounded yearly. How much money is in the account after 7 years (assu

julia-pushkina [17]

Answer:

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Step-by-step explanation: You can use the formula for compound interest:

P(1+R)^t

P = the principle, the amount of money

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^t = to the power of the number of years

Substitute the values:

750(1+0.08)^7

750(1.08)^7

1285.37

8 0

4 years ago

Read 2 more answers