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4 years ago

(0,0) and (2,3) slope

Mathematics

1 answer:

Oksanka [162]4 years ago

6 0

Answer:

From (0,0) and (2,3) the slope is 2/3.

Step-by-step explanation:

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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

Is f(x) continuous at x equals 4? Why or why not? A. No, f(x) is not continuous at x equals 4 because ModifyingBelow lim Wit

soldier1979 [14.2K]

Corrected Question

Is the function given by:

$f(x)=\left\{\begin{array}{ccc}\frac{1}{4}x+1 &x\leq 4\\4x-11&x>4\end{array}\right$

continuous at x=4? Why or why not? Choose the correct answer below.

Answer:

(D) Yes, f(x) is continuous at x = 4 because $Lim_{x \to 4}f(x)=f(4)$

Step-by-step explanation:

Given the function:

$f(x)=\left\{\begin{array}{ccc}\frac{1}{4}x+1 &x\leq 4\\4x-11&x>4\end{array}\right$

A function to be continuous at some value c in its domain if the following condition holds:

f(c) exists and is defined.

$Lim_{x \to c}$ f(x)$ exists.

$f(c)=Lim_{x \to c}$ f(x)$

At x=4

$f(4)=\dfrac{1}{4}*4+1=2$

$Lim_{x \to 4}f(x)=2$

Therefore: $Lim_{x \to 4}f(x)=f(4)=2$

By the above, the function satisfies the condition for continuity.

The correct option is D.

3 0

3 years ago

Label the parts of the triangle. leg leg altitude hypotenuse right angle

andriy [413]

Answer: See Below

Step-by-step explanation:

One part we can label right off the bat is right angle. It is denoted with the red box. It forms a 90° angle, which is a right angle. At the very top pointing to the right angle, is right angle.

Altitude is another way of saying height. It is how tall the triangle is. The line that shows how tall the triangle is, is the line going down the triangle in the middle. That line is altitude.

Hypotenuse is the longest side of the triangle. It is also located directly across the right angle. Looking at the right angle, the bottom line is the hypotenuse.

Since we have 2 sides left, they are obviously the legs of the triangle.

6 0

3 years ago

Where would (-3,2) be if it was reflected across the x-axis

Furkat [3]

Answer:

(-3,-2)

Step-by-step explanation:

When you reflect over the x-axis, the x remains the same but the y changes to the opposite. In this case negative. I hope this helps, have a great day!

4 0

3 years ago

Q(x) = 2x+5 f(x)= x^2+1 find (gof)(7)

poizon [28]

Answer:105

Step-by-step explanation:

first find f(7)

(7)^2 + 1

=50

then put the answer for f(7) in g of x

2(50) + 5

=105

4 0

4 years ago