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4 years ago

9

point C(4,2) divides the line segment joining points A(2,-1) and B(x,y), such that AC:CB=3.1. what are the coordinates of point

B?

1 answer:

White raven [17]4 years ago

6 0

Answer: $B(x,y)=B(\frac{14}{3},3)$

Step-by-step explanation:

$\frac{4-2}{3} =x-4\\2=3x-12\\14=3x\\\frac{14}{3}=x\\\\\frac{2-(-1)}{3}=y-2\\3=3y-6\\9=3y\\3=y$

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Sophie was 28 years old when her daughter Nichole was born. In how many years will Sophie be two years less than six times Nicho

artcher [175]

It will take 6 years for that to happen.

You can find this by setting Nichole's age as x and Sophie's age as x + 28 since she will always be 28 years older.

Now we are looking for a time that Sophie's age is equal to 2 less than 6 times Nichole's age. So we can set up the below equation.

6(x) - 2 = x + 28
6x - 2 = x + 28
5x - 2 = 28
5x = 30
x = 6

This let's us know that it needs to be when Nichole is 6, which is 6 years in the future.

5 0

4 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

Write an equation in standard form of the line that is

katen-ka-za [31]

What is boxed is what you put in your stat plot in your calculator (graphing). To find the y intercept you use the value function on your calculator (if you have a Ti-84 it’s 2nd-> trace-> 1) and enter 0. Yo find the x intercept you use the zero function on your calculator (if you have a Ti-84 it’s 2nd-> trace-> 2 and follow the instructions it asks). Hope this helps!

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3 years ago

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What is the value of b in the equation below?

jarptica [38.1K]

when dividing subtract the powers

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4 years ago

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