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That is right! That is what I got by having the the computer arrange them in order and also make sure that all are divisible by 4. You have to have the ability to program the computer in some language such as Python, Java, C.......etc. Here is the list:

(2456, 2476, 2564, 2576, 2756, 2764, 4256, 4276, 4572, 4576, 4652, 4672, 4752, 4756, 5264, 5276, 5472, 5476, 5624, 5672, 5724, 5764, 6452, 6472, 6524, 6572, 6724, 6752, 7256, 7264, 7452, 7456, 7524, 7564, 7624, 7652) = 36 - numbers and all are divisible by 4.

Suppose that the greatest common divisor is k, and that

\(\displaystyle 13n+8=m_{1}k\dots (1)\\\text{and that }\\5n+3=m_{2}k\dots(2).\)

From (1)

\(\displaystyle 65n+40=5m_{1}k\dots(3) \\ \text{and from (2),} \\ 65n+39=13m_{2}k\dots(4).\)

Subtracting (4) from (3), 

\(\displaystyle 1=(5m_{1}-13m_{2})k,\)

implying that k = 1.

I've never seen a rooster that can lay eggs......perhaps it is a hen that IDENTIFIES as a rooster?   Haha  ! 

Thanks 

I will assume you mean the elevation in relation to sea level....

  I do not know why they gave you City A stats....sometimes Q's give information to mislead ya or to make sure you know what you are doing....

Part A       D_level = 1/4  C_level

            = 1/4 (-24) = -6 ft   which is the same as saying    = 6 ft BELOW sea level  (this is Part B)

Temp in city for weeks 1-4 =  Starting temp - (20 deg C / 4 wk) * wk#   

Temp in City for weeks 1-4 = starting temp - 5 deg c/wk  * wk#

This shows the temperature in the city for weeks 1 2 3 4  given the original temperature of the city and whichever week you want to know from the starting point....week 1   week 2   week 3   week 4 .....

How are you able to find the bearing?

Let n be a positive integer greater than or equal to 3.
Let a,b be integers such that is invertible modulo n and \((ab)^{-1}\equiv 2\pmod n\).
Given a+b is invertible, what is the remainder when \((a+b)^{-1}(a^{-1}+b^{-1})\) is divided by n?

\(\begin{array}{|rcll|} \hline (ab)^{-1} &\equiv& 2\pmod n \\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot a \\ \mathbf{b^{-1}} &\equiv&\mathbf{ 2a\pmod n} \\\\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot b \\ \mathbf{a^{-1}} &\equiv& \mathbf{2b\pmod n} \\\\ (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& x \pmod n \\ (a+b)^{-1}(2b+2a) &\equiv& x \pmod n \\ 2(a+b)^{-1}(ab) &\equiv& x \pmod n \\ 2 &\equiv& x \pmod n \\ x &\equiv& 2 \pmod n \\ \hline \end{array}\)

What is the largest integer \(n\) such that \(7^n\) divides \(1000!\)

\(\text{The largest integer $= \left[\dfrac{1000}{7}\right] + \left[\dfrac{1000}{7^2}\right] + \left[\dfrac{1000}{7^3}\right] \qquad [\ldots] =$ integer part. } \\ \text{The largest integer $= 142 + 20 + 2 $} \\ \text{The largest integer $= 164 $} \)

1.
What is the smallest distance between the origin and a point on the graph of \(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)?\)

\(\text{Let the distance $=s$} \)

\(\begin{array}{|rcll|} \hline s^2 &=& x^2+y^2 \quad &|\quad y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right) \\ s^2 &=& x^2+\left(\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\right)^2 \\ s^2 &=& x^2+\dfrac{1}{2}\left(x^2-3\right)^2 \\ \dfrac{d\ s^2}{dx} &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}(1+x_{\text{min}}^2-3) \\ &=& 2x_{\text{min}}(x_{\text{min}}^2-2) \\\\ \dfrac{d\ s^2}{dx} &=& 0 \\ 2x_{\text{min}}(x_{\text{min}}^2-2) &=& 0 \\ x_{\text{min}}^2-2 &=& 0 \\ \mathbf{ x_{\text{min}}^2 } &=& \mathbf{2} \\\\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(x_{\text{min}}^2-3\right) \\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(2-3\right) \\ \mathbf{ y_{\text{min}} } &=& -\dfrac{1}{\sqrt{2}} \\\\ s_{\text{min}} &=& \sqrt{x_{\text{min}}^2+y_{\text{min}}^2} \\ s_{\text{min}} &=& \sqrt{2+\left(-\dfrac{1}{\sqrt{2}}\right)^2} \\ s_{\text{min}} &=& \sqrt{ 2+ \dfrac{1}{2} } \\ s_{\text{min}} &=& \sqrt{ \dfrac{5}{2} } \\ \mathbf{ s_{\text{min}}} &=& \mathbf{1.58113883008\ldots} \\ \hline \end{array}\)

Suppose that x is an integer that satisfies the following congruences:
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
What is the remainder when x is divided by 105?

Great Solution, Heureka!

idk how. can u like, explain how this can be done easily?

As n  ranges over the positive integers, what is the maximum possible value that the greatest common divisor of  13n + 8 and  5n+3 can take?

Formula: \(\boxed{\gcd(a,b)=gcd(a-b,b)}\)

\(\begin{array}{|rcll|} \hline && \mathbf{gcd(13n + 8, 5n+3)} \\ &=& gcd\left(13n+8-(5n+3), 5n+3\right) \\ &=& gcd\left(13n+8-5n-3, 5n+3\right) \\ &=& gcd\left(8n+5, 5n+3\right) \\ &=& gcd\left(8n+5-(5n+3), 5n+3\right) \\ &=& gcd\left(8n+5-5n-3, 5n+3\right) \\ &=& gcd\left(3n+2, 5n+3\right) \\ &=& gcd\left(5n+3,3n+2\right) \\ &=& gcd\left(5n+3-(3n+2),3n+2\right) \\ &=& gcd\left(5n+3-3n-2,3n+2\right) \\ &=& gcd\left(2n+1,3n+2\right) \\ &=& gcd\left(3n+2,2n+1\right) \\ &=& gcd\left(3n+2-(2n+1),2n+1\right) \\ &=& gcd\left(3n+2-2n-1,2n+1\right) \\ &=& gcd\left(n+1,2n+1\right) \\ &=& gcd\left(2n+1,n+1\right) \\ &=& gcd\left(2n+1-(n+1),n+1\right) \\ &=& gcd\left(2n+1-n-1,n+1\right) \\ &=& gcd\left(n,n+1\right) \\ &=& gcd\left(n+1,n\right) \\ &=& gcd\left(n+1-(n),n\right) \\ &=& gcd\left(n+1-n,n\right) \\ &=& gcd\left(1,n\right) \\ &=& gcd\left(n,1\right) \\ &=& \mathbf{1} \\ \hline \end{array}\)

The maximum possible value is \(\mathbf{1}\)

The circle centered at \((2,-1)\) and with radius \(4\) intersects the circle centered at \((2,5)\) and with radius \(\sqrt{10}\) at two points \(A\) and \(B\).
Find \(\left(\overline{AB}\right)^2\).

\(\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \\ & (x-2)^2 +(y-(-1))^2 &=& 4^2 \\ (2) & (x-2)^2 +(y+1)^2 &=& 16 \\ \hline (2)-(1): & (x-2)^2 +(y+1)^2 -((x-2)^2 +(y-5)^2) &=& 16-10 \\ & (x-2)^2 +(y+1)^2 -(x-2)^2 -(y-5)^2 &=& 6 \\ & (y+1)^2 -(y-5)^2 &=& 6 \\ & y^2+2y+1-(y^2-10y+25) &=& 6 \\ & y^2+2y+1-y^2+10y-25 &=& 6 \\ & 12y+1-25 &=& 6 \\ & 12y-24 &=& 6 \quad &| \quad : 6 \\ & 2y-4 &=& 1 \\ & 2y &=& 1+4 \\ & 2y &=& 5 \\ & \mathbf{ y } &=& \mathbf{\dfrac{5}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \quad | \quad y=\dfrac{5}{2} \\ & (x-2)^2 +\left(\dfrac{5}{2}-5\right)^2 &=& 10 \\ & (x-2)^2 +\left(-\dfrac{5}{2}\right)^2 &=& 10 \\ & (x-2)^2 + \dfrac{25}{4} &=& 10 \\ & (x-2)^2 &=& 10 - \dfrac{25}{4} \\ & (x-2)^2 &=& \dfrac{15}{4} \\ & x-2 &=& \pm\dfrac{\sqrt{15}}{2} \\ & x &=& 2 \pm \dfrac{\sqrt{15}}{2} \\ \\ & \mathbf{x_1} &=& 2 + \dfrac{\sqrt{15}}{2} \\ & \mathbf{x_2} &=& 2 - \dfrac{\sqrt{15}}{2} \\\\ & \mathbf{ \left(\overline{AB}\right)^2} &=& (x_1-x_2)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}- \left(2 - \dfrac{\sqrt{15}}{2}\right) \right)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}-2 + \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\dfrac{\sqrt{15}}{2}+ \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(2\dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\sqrt{15}\right)^2 \\ & &=& \mathbf{15} \\ \hline \end{array}\)

What's a conjugate again?

It's fine.

Points\( A\) and \(B\) are on parabola \(y=3x^2-5x-3\), and the origin is the midpoint of \(\overline{AB}\).
Find the square of the length of \(\overline{AB}\).

\(\text{ Let $A(x_A,\ y_A)$ and $B(x_B,\ y_B)$ }\)

\(\begin{array}{|lrcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ (2) & y_B&=&3x_B^2-5x_B-3 \\ (3) & x_B&=&-x_A\\ (4) & y_B&=&-y_A\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ \hline (2) & y_B&=&3x_B^2-5x_B-3 \quad &| \quad y_B=-y_A,\ x_B=-x_A \\ & -y_A&=&3(-x_A)^2-5(-x_A)-3 \\ (2)^* & -y_A&=&3x_A^2+5x_A-3 \\ \hline (1)+(2)^*: & y_A-y_A&=&3x_A^2-5x_A-3+3x_A^2+5x_A-3\\ & 0&=&6x_A^2-6 \quad &| \quad : 6 \\ & 0&=&x_A^2-1 \\ & x_A^2 &=& 1 \\ & \mathbf{x_A} &=& \mathbf{1} \\ & \mathbf{x_B} &=& -x_A = \mathbf{-1} \\ \hline & y_A &=& 3x_A^2-5x_A-3 \\ & &=& 3(1)^2-5(1)-3 \\ & &=& 3-5-3 \\ & \mathbf{y_A} &=& \mathbf{-5} \\ & \mathbf{y_B} &=& -y_A = -(-5) = \mathbf{5} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A(1,\ -5),\ B(-1,\ 5) \\\\ \text{The square of the length of } \overline{AB} : \\ \left(1-(-1)\right)^2 + \left(-5-(5)\right)^2\\ =2^2+(-10)^2 \\ =4+100 \\ \mathbf{= 104} \\ \hline \end{array} \)

Are you putting hw probs here so other people can answer it and so you can pass your class?

Just curious.

You think?

\(\text{you can see the zero crossings are at $x=0, x=1, x=2$}\\ \text{Thus the curve is given by }\\ f(x) = x(x-1)(x-2) = x^3-3 x^2+2 x\\ a+b+c+d =1-3+2= 0\)

\(\text{one can be a bit more clever and note that in general}\\ a+b+c+d = f(1)\\ \text{and we know that $1$ is one of the zeros so}\\ a+b+c+d=f(1)=0\)

We need to find how many factors of seven all the integers from 1 to 1000 have.

Each integer from 1 to 1000 can be a factor of 7, 49, or 343, giving it 1 factor of 7, 2 factors of 7, or 3 factors of 7. 

There are \(\lfloor\frac{1000}{7}\rfloor=142\) numbers that have 1 factor of 7.

There are \(\lfloor\frac{1000}{49}\rfloor=20\) numbers that have 2 factors of 7.

There are \(\lfloor\frac{1000}{343}\rfloor=2\) numbers that have 3 factors of 7.

So, the answer is \(2\cdot3+20\cdot2+142\cdot1\).

But, wait!

We are counting numbers that have 2 factors in the numbers that have 1 factor of 7!

And, we are counting numbers that have 3 factors in the numbers that have 2 factors and the numbers that have 1 factor!

So, the answer is \(2\cdot3+20\cdot2+142\cdot1-2-2-20=2+20+142=164\)!

Thanks!

Ooops.... forgot to log in (again) !      ~ EP

The x component will be given by

magnitude * cos (theta) = x component

so

Magnitude = (x component) / (cos theta)

The boat travels directly across the river at a velocity......this component of its 'final' or resultant velocity is perpindicular to 'v'  .

   the boat ALSO travels downriver at velocity 'v' ....this component is perpindicular to its crossing velocity, but it is a component of its resultant velocity.

Another way:

  The boat has two perpindicular  components of its 'resultant' velocity..... the downriver current, v , and its crossing velocity (provided by its motor...or sail...or paddle etc).