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4. Given

\(a \equiv 1 \pmod{7}\),

\(b \equiv 2 \pmod{7}\),

and

\(c \equiv 6 \pmod{7}\),

what is the remainder when

\(a^{81} b^{91} c^{27} \)

is divided by 7?

\(\begin{array}{|rclcl|} \hline a-1 &=& 7l \quad &\text{or}& \quad a =7l+1 \\ b-2 &=& 7m \quad &\text{or}& \quad b =7m+2 \\ c-6 &=& 7n \quad &\text{or}& \quad c =7n+6 \\ \hline \end{array}\)

\(\begin{array}{|rclcl|} \hline && a^{81} b^{91} c^{27} \pmod 7 \\ &\equiv& (7l+1)^{81}\cdot (7m+2)^{91}\cdot (7n+6)^{27} \pmod 7 \\\\ &\equiv& \left[\binom{81}{0}(7l)^{81}+ \binom{81}{1}(7l)^{80} +\dots + \binom{81}{81} 1^{81} \right] \\ && \cdot \left[\binom{91}{0}(7m)^{91}+ \binom{91}{1}(7m)^{90} +\dots + \binom{91}{91} 2^{91} \right] \\ && \cdot \left[\binom{27}{0}(7n)^{27}+ \binom{27}{1}(7n)^{26} +\dots + \binom{27}{27} 6^{27} \right] \pmod 7 \\\\ &\equiv& \binom{81}{81} 1^{81}\cdot \binom{91}{91} 2^{91} \cdot \binom{27}{27} 6^{27}\pmod 7 \\\\ &\equiv& 1^{81}\cdot 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 1 \cdot 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 2^{91} \cdot 6^{27}\pmod 7 \\ \hline \end{array} \)

\(\begin{array}{|lllcl|} \hline \text{Because } gcd(2,7) = 1 \quad \Rightarrow & 2^{\varphi(7)} \equiv 1 \pmod 7 \qquad \varphi(7) = 6\\ & \mathbf{2^{6} \equiv 1 \pmod 7} \\\\ \text{Because } gcd(6,7) = 1 \quad \Rightarrow & 6^{\varphi(7)} \equiv 1 \pmod 7 \qquad \varphi(7) = 6\\ & \mathbf{6^{6} \equiv 1 \pmod 7} \\ \hline \end{array} \)

\(\begin{array}{|rclcl|} \hline && a^{81} b^{91} c^{27} \pmod 7 \\ &\equiv& 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 2^{6\cdot 15+1} \cdot 6^{6\cdot 4+3}\pmod 7 \\ &\equiv& 2^{6\cdot 15}\cdot 2 \cdot 6^{6\cdot 4}\cdot 6^3 \pmod 7 \\ &\equiv& (2^6)^{15}\cdot 2 \cdot (6^6)^{4}\cdot 6^3\pmod 7 \qquad \mathbf{2^{6} \equiv 1 \pmod 7}\\ &\equiv& (1)^{15}\cdot 2 \cdot (6^6)^{4}\cdot 6^3\pmod 7 \qquad \mathbf{6^{6} \equiv 1 \pmod 7}\\ &\equiv& (1)^{15}\cdot 2 \cdot (1)^{4}\cdot 6^3\pmod 7 \\ &\equiv& 1 \cdot 2 \cdot 1 \cdot 6^3\pmod 7 \\ &\equiv& 2 \cdot 6^3 \pmod 7 \\ &\equiv& 2 \cdot 216 \pmod 7 \\ &\equiv& 432 \pmod 7 \\ &\equiv& \mathbf{5} \pmod 7 \\ \hline \end{array} \)

The remainder is 5

How about this?????? in words ......10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

One Hundred Quintillion

1. Let  \(a \equiv 1 \pmod{4}\)

Find the value of 6a + 5 (mod 4). Express your answer as a residue between 0 and the modulus.

6a + 5 (mod 4)

=(6*1 + 5) (mod 4)

=11(mod4)            

            11/4=2R3      or      3R-1

= 3 or -1  (mod4)

I'm trying to get this to work for all values of m.

https://www.desmos.com/calculator/ars8nrqtyw

Any idea's?

The line  y = tan(pi/4)·x  intersects the circle  x² + y² =  1

at the points  ( -sqrt(2)/2, -sqrt(2)/2 )  and  ( sqrt(2)/2, sqrt(2)/2 ).

In general:  the x-values of the points of intersection of the circle x² + y² =  1 and y = mx can be found by

replacing the value of y in the equation of the circle with the value mx   --->   x² + (mx)² =  1

      --->   x² + m²x² =  1     --->   x²(1 + m²)  =  1     --->   x²  =  1 / (1 + m²)

      --->   lower value:  x  =  - sqrt( 1 / (1 + m²) )

      --->   upper value:  x  = sqrt( 1 / (1 + m²) )

A statistic refers to a sample; a parameter refer to a population.

Since the reference is to the unemployment rate for the USA (a population), it is a parameter.

One other question...

What would the range of this function be, if you wanted to limit the length of the segment to 'fit' inside the circle.

\(x^2+y^2=1\)

\(y=\tan \left(\frac{\pi }{4}\right)x\)

It represents an inference about teenagers in the United States.

You are generalizing from the 33%  from the random sample to the population at large.

Of course... thank you :)

The slope of a line is equal to the tangent of the angle that is formed by the line with the positive x-axis.

Thus, the slope of a line at an angle of pi/4 radians = tan(pi/4) = 1.

To factor  12x² - 6x - 90, it would be OK to start by factoring out 3:

--->   12x² - 6x - 90  =  3(4x² - 2x - 30)

This reduces the number of options without removing any possibly correct factors.

Also, if you look at  4x² - 2x - 30  you may notice that each of the coefficients is an even number, meaning that you can now factor out a 2, making the problem simpler.

By factoring out both a 3 and a 2, you are factoring out a 6:

--->   12x² - 6x - 90  =  6(2x² - x - 15)

leaving you with far fewer possibilities.

Wait, but doesn't ((137+133)/2)^(1/3) simplify into (270/2)^(1/3) which is (135/2)^(1/3), not (125/2)^1/3. 

Let the mystery number =N

{[N+38] x 4 - 26,729} =209,423

Solve for N:
4 (N+38)-26729 = 209423

4 (N+38) = 4 N+152:
4 N+152-26729 = 209423

Add like terms. 152-26729 = -26577:
4 N-26577 = 209423

Add 26577 to both sides:
4 N+(26577-26577) = 209423+26577

26577-26577 = 0:
4 N = 209423+26577

209423+26577 = 236000:
4 N = 236000

Divide both sides of 4 N = 236000 by 4:
(4 N)/4 = 236000/4

4/4 = 1:
N = 236000/4

The gcd of 236000 and 4 is 4, so 236000/4 = (4×59000)/(4×1) = 4/4×59000 = 59000:
Answer: |N = 59,000

Since the track is 5 3/5 feet long, he will need 5 poles, at the end of 1st, 2nd,3rd,4th and 5th foot.

One general formula for a vertex is:  y - k  =  a(x - h)²     <--->     This parabola has its vertex at (h,k).

If a parabola has a vertex (5,3), its equation can be:  y - 3  =  a(x - 5)²

To find the value of a, place (2,0) into this equation, replacing x with 2 and y with 0:   0 - 3  =  a(2 - 5)²

and solve this for a:     0 - 3  =  a(2 - 5)²     --->     -3  =  a(-3)²     --->     -3  =  a·9     --->    a  =  -1/3

So, the equation is:  y - 3  =  (-1/3)(x - 5)²

Multiplying this out:     y - 3  =  (-1/3)(x² - 10x + 25)

                                   y - 3  =  (-1/3)x² + (10/3)x - (25/3)

                                        y  =  (-1/3)x² + (10/3)x - (16/3)

--->     a = -1/3     b = 10/3     c = -16/3

--->     a + b + c  =  -7/3

log base n (x) = log base a (b)

X=?

X=n^(log (b) / log(a))

Example: Log(base 2) of X =Log(base 5) of 100

X=2^(log(100)/log(5)) =7.266965.......etc.

Since  log_n(x) = y  can be written as  n^y  =  x  or  x = n^y,

log_n(x)  =  log_a(b)   can be written as   n^log_a(b)  =  x      or     x  =  n^log_a(b)​  

Hello Guest!

\(subtract (3/9x^2-5/6x-14) from (7/9x^2-5/6x+1)\)

\({(\frac{7}{9x^{2} }- \frac{5}{6x}+ 1 )} -{(\frac{3}{9x^{2} }- \frac{5}{6x} - 14) }\)

\(= \frac{7- 3}{9x^{2} }+ 1+ 14 = \frac{4}{9x^{2} }+15 = (\frac{2}{3x}) ^{2} + 15 \)

Greeting asinus :- )  !

6 x 7 x 8 x 9 =3,024

6+7+8+9 =30

What is x^4y^3 ·2y^1?

=2x^4y^4

Of course, it does! Press "2nd" key in the top left hand corner. You should see "atan", which is the same as arctangent. Press "=" key to get your answer.

Simplify the following:
-((3 x^2)/9-5/6 x-14)+7/9 x^2-5/6 x+1

3/9 = 3/(3×3) = 1/3:
-(x^2/3-5/6 x-14)+7/9 x^2-5/6 x+1

Put each term in x^2/3-5/6 x-14 over the common denominator 6: x^2/3-5/6 x-14 = (2 x^2)/6-(5 x)/6-84/6:
-(2 x^2)/6-(5 x)/6-84/6+7/9 x^2-5/6 x+1

(2 x^2)/6-(5 x)/6-84/6 = (2 x^2-5 x-84)/6:
(7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6

Put each term in (7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6 over the common denominator 18: (7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6 = (14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18:
(14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18

(14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18 = (14 x^2-15 x+18+3 (-2 x^2+5 x+84))/18:
(14 x^2-15 x+18+3 (-2 x^2+5 x+84))/18

3 (-2 x^2+5 x+84) = -6 x^2+15 x+252:
(-6 x^2+15 x+252+14 x^2-15 x+18)/18

Grouping like terms, 14 x^2-6 x^2+15 x-15 x+18+252 = (14 x^2-6 x^2)+(-15 x+15 x)+(18+252):
((14 x^2-6 x^2)+(-15 x+15 x)+(18+252))/18

14 x^2-6 x^2 = 8 x^2:
(8 x^2+(-15 x+15 x)+(18+252))/18

18+252 = 270:
(8 x^2+(-15 x+15 x)+270)/18

15 x-15 x = 0:
(8 x^2+270)/18

Factor 2 out of 8 x^2+270:
2 (4 x^2+135)/18

2/18 = 2/(2×9) = 1/9:
Answer: |(4x^2+135) / 9

Hello bla!

During a space shuttle launch, a maneuver is scheduled to begin

at T minus 80 seconds, which is 80 seconds before liftoff.

The maneuver lasts 1 1/2 minutes.

At what time will the maneuver be complete?

\(-80sec+ 1\ \frac{1}{2}\min \times \frac{60sec}{min} =-80sec+ 90sec= 10sec\) 

The maneuver is completed 10 seconds

after lift-off of the projectile.

Greeting asinus :- )

  !

I need an answer to this question  --_--