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3 years ago

14

For each statement below enter a T (true) the statement is true and an F (false) otherwise. In this problem you need to get ever

ything correct before receiving credit. Whenever there is a division below, we assume that the divisor is non-zero. We also assume that where it matters the base of any power is positive.
_____For all real numbers a, b, and c a^b a^c = a^be
_____For all real numbers a, b, and c a^b a^c = a^b+c
_____For all real numbers a and b a^-b= a/a^b
_____For all real numbers a and b a^-b = -a^b
_____For all real numbers a and b a^b = ab

1 answer:

aleksley [76]3 years ago

3 0

Answer:

T

T

T

T

T

T

F

Step-by-step explanation:

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What's the answer to 22+4y-14=0

Gwar [14]

Y=-2 You combine like terms which would be -14 and 22. 22-14=8. Then you subtract that from 0. 0-8=-8. Then divide -8 by 4y. -8÷4=-2

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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

A rectangular swimming pool measures 14 feet by 30 feet. The pool is surrounded on all four sides by a path that is 3 feet wide.

V125BC [204]

Check the picture below.

since the rectangular pool is a 14x30, the top and bottom part of that rectangle in the picture are just a 3x30 piece and the sides are 3x8, so how many ft² is that?

$\bf 2(3\cdot 30)+2(3\cdot 8)\implies 180+48\implies 228~ft^2~\hfill \stackrel{\textit{total cost}}{228\cdot 20\implies \boxed{456}}$

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4 years ago

Read 2 more answers

Does anyone know the answer to the questions above thank you!

dolphi86 [110]

2 would be "B" and 3 would "D"

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4 years ago

How do you solve linear equations using substitution

GuDViN [60]

You substitute the y from the second eqaution to the y in the first equation

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3 years ago