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3 years ago

Can anyone help me with Combining like terms with negative coefficients I'm not good at math

Mathematics

1 answer:

Viktor [21]3 years ago

7 0

Yes if you give me brainiest

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(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) 169p⁴qr÷ (-13pq³r)<br> b) 3(x +2y) +5x - (y + 7)<br> c (3x –4)( 5x- 1)

Veseljchak [2.6K]

Answer:

A) -13p^33q^-2 or -13p^3/q^2

B) 8x + 5y - 7

C) 15x^2 -23x + 4

Step-by-step explanation:

A) 169/(-13) = -13 p^4/p = p^3 q/q^3 = 1/q^2 which can be written as q^-2 and r/r is 1. So if you put this all together you get -13p^33q^-2 or -13p^3/q^2

B) 3(x +2y) + 5x - (y + 7)

3x + 6y + 5x - y - 7 distributive property

3x + 5x + 6y - y -7 combine like terms

8x + 5y - 7 Answer

C) (3x-4) (5x - 1) Foil multiply the first terms in each parenthesis. the outer terms, the inner terms and the last terms

(3x)(5x) + (3x)(-1) + (-4)(5x) + (-4)(-1)

15x^2 -3x - 20x + 4

15x^2 -23x + 4

5 0

3 years ago

Please help brainiest is award.

lesantik [10]

it's easier b or d I'm not sure

8 0

3 years ago

Read 2 more answers

Please solve the following quadratic equation and show your work: x^2-2x+6=0

hichkok12 [17]

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ exist\ two\ solutions\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta =0\ then\ exist\ one\ solution\ x_o=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solutions$

We have:

$x^2-2x+6=0\\\\a=1;\ b=-2;\ c=6\\\\\Delta=(-2)^2-4\cdot1\cdot6=4-24=-20 < 0$

Answer: NO SOLUTIONS.

6 0

4 years ago

The function f(x) varies directly with x and f(x)=90 when x = 30. What is f(x) when x = 6?

maria [59]

Answer: just took the test f(x) = 18

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers