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Hello Guest! I suppose you are one of the people who attend Middle School Competition Math! I would kindly advise not to post the questions on here, unless you want to be scolded by the director! Thank you, and have a good day.

ln(e) = 1 so it needn't be there, (shouldn't be there).  It looks like it's a ' feature ' of your calculator.

In the middle line, the first derivative is wrt ax, the second simply wrt x, it isn't that you are simply replacing a by 3.

On the third line, study the effect of replacing ax by the variable y (=ax),  ie. d/dy(e^(y) )= e^(y).

Thank you very much!

What you are seeing is Rom’s resignation –retroactive to the late hours of the prior week.

Last two lines heureka, log is base 10.

Hello, if i want the Differential from "log10(ax)" what happens to the a ?
i would guess "1/a*x*ln(10)" but i am not sure because in the diff from "ln(ax)" the a fades away because "a*1/a*X"

\(\begin{array}{|rcll|} \hline y &=& \log_{10}(ax) \\ y &=& \log_{10}(a)+\log_{10}(x) \\ \hline y' &=& \underbrace{\dfrac{d\ \log_{10}(a) }{dx}}_{=0} + \dfrac{d\ \log_{10}(x) }{dx} \\\\ \mathbf{y'} &=& \mathbf{\dfrac{d\ \log_{10}(x) }{dx}} \\\\ \mathbf{y'} &=& \mathbf{\dfrac{1}{x\ln(10)}} \\ \hline \end{array}\)

Corrected. Thank you Guest!

The following points are the vertices of a parallelogram. (0,0), (2,3), (7,3), (5,0)

Write a system of linear inequalities that represent the parallelogram.

Find \(a\)  such that \(ax^2+12x+9\) is the square of a binomial.

\(\begin{array}{|rcll|} \hline ax^2+12x+9 &=& a(x-x_0)^2 \\\\ ax^2+12x+9 &=& a(x^2-2xx_0+x_0^2) \\ ax^2+12x+9 &=& ax^2-2xx_0a+x_0^2a \\ 12x+9 &=& -2xx_0a+x_0^2a \quad | \quad \text{compare: } 12 = -2x_0a,\ 9 = x_0^2a \\ \hline && \begin{array}{|rcll|} \hline 12 &=& -2x_0a \\ 6 &=& -x_0a \\ \mathbf{x_0} &=& \mathbf{- \dfrac{6}{a} } \\\\ 9 &=& x_0^2a \\ 9 &=& \left(- \dfrac{6}{a}\right)^2a \\ 9 &=& \dfrac{36a}{a^2} \\ 9 &=& \dfrac{36 }{a } \\ a &=& \dfrac{36 }{9 } \\ \mathbf{a} &=& \mathbf{4} \\ \hline \end{array} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x_0 &=& -\dfrac{6}{a} \quad | \quad a = 4 \\\\ x_0 &=& -\dfrac{6}{4} \\\\ \mathbf{x_0} &=& \mathbf{-\dfrac{3}{2} } \\ \hline && a(x-x_0)^2 \\ &=& 4\left(x+\dfrac{3}{2}\right)^2 \\ &=& \left(2\left(x+\dfrac{3}{2}\right)\right)^2 \\ &=& \left( 2x+2\cdot\dfrac{3}{2} \right)^2 \\ &=& ( 2x+3 )^2 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{4x^2+12x+9} &=& \mathbf{( 2x+3 )^2} \\ \hline \end{array}\)

I would substitute \(z=(x+y)e^x \)

Then you get \(\frac{dz}{dx}=z^2e^x\)  which is easy to solve by separation of variables.

Thanks Tiggsy that thought had crossed my mind too but I did not attempt to think it through.

2)
Grogg skip counts by 11s starting at 12. He obtains the sequence: \(12, 23, 34, \ldots, 1233\).
What is the remainder when the sum of all the terms in Grogg's sequence is divided by 11?

\(\begin{array}{|rcll|} \hline &&\mathbf{ 12 + \underbrace{(12+11)}_{23} + \underbrace{(12+2\cdot 11)}_{34}+ \ldots + \underbrace{(12+(n-1)\cdot 11)}_{1233} \pmod{11} }\\ &\equiv& 12 +(12+0) + (12+0)+ \ldots + (12+0) \pmod{11} \\ &\equiv& 12n \pmod{11} \quad | \quad 12\pmod{11} = 1 \\ &\equiv& 1n \pmod{11} \\ &\equiv& n \pmod{11} \\ && \boxed{ 12+(n-1)\cdot 11 = 1233\\ 12+11n-11 =1233\\ 1+11n = 1233\\ 11n = 1232\\ n=\dfrac{1232}{11}\\ n = 112 } \\ &\equiv& 112 \pmod{11} \\ &\equiv& 110 + 2 \pmod{11} \quad | \quad 110 \pmod{11} = 0 \\ &\equiv& 0 + 2 \pmod{11} \\ &\equiv& \mathbf{ 2 \pmod{11} } \\ \hline \end{array} \)

The remainder when the sum of all the terms in Grogg's sequence is divided by 11 is 2

My guess, Guests, and Melody, is that it's the area of the triangle.

I know pretty much nothing about critical path analysis so this is simply a common sense view of the situation.

The problem seems to be with the join of C and D and specifically F from that point.

F depends upon D but not upon C, so activity F can be started as soon as D is finished and before C is finished.

That means you have to separate that join of C and D in some way. 

I don't know how, conventionally,  this is represented.

Thanks guest, maybe you are right.     

1)
Suppose n is a positive number between 1 and 11 inclusive, and \(n^2 \equiv 0 \pmod{12}\).
What is \(n\)?

\(\begin{array}{|rcll|} \hline n^2 &\equiv & 0 \pmod{12} \quad | \quad 0 \equiv 12 \equiv 24 \equiv 36 \pmod{12} \quad 36\text{ is a perfect square}\\ n^2 &\equiv & 36 \pmod{12} \quad | \quad sqrt() \\ n &\equiv & \pm \sqrt{36} \pmod{12} \\ \hline && \begin{array}{|rclcrcl|} \hline n &\equiv & 6 \pmod{12} &or & n &\equiv & -6 \pmod{12} \\ & & && n &\equiv & -6+12 \pmod{12} \\ & & && n &\equiv & 6 \pmod{12} \\ \hline \end{array} \\ \hline \end{array}\)

\(n\) is 6

Your first step should be the substitution z = x + y.

That gets you the equation

\(\displaystyle \frac{dz}{dx}+ z=z^{2}e^{2x}. \)

Now use the usual Bernoulli substitution.

Calculate the root-mean-square velocity and kinetic energy of CO, CO2, and SO3 at 282 K Part A Calculate the root-mean-square velocity of CO at 282 K .

Piecewise function

If  \(z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \)  find \(\lfloor z \rfloor\).

The fractional part denoted by  \(\{x\}\) for real x and defined by the formula \( \{x\}=x-\lfloor x\rfloor\).

Suppose you want to fill the 5x5 square below with the following restrictions:
(1) The numbers 1-5 must appear in every row and column;
(2) The sum of the numbers in gray boxes is 2 (mod 6).

What number must go in the center square?

Let the number go in the center square \(= \mathbf{x}\)

Let the sum of the numbers \(1+2+3+4+5 = 15\)

Let the sum of the numbers in gray boxes \(= 15 + (15-x)\)

Then:

\(\begin{array}{|rcll|} \hline 15 + (15-x) & \equiv & 2 \pmod 6 \\ 30-x & \equiv & 2 \pmod 6 \quad | \quad 30 \pmod 6 = 0 \\ 0-x & \equiv & 2 \pmod 6 \\ -x & \equiv & 2 \pmod 6 \quad | \quad \cdot (-1) \\ x & \equiv & -2 \pmod 6 \\ x & \equiv & -2+6 \pmod 6 \\ x & \equiv & 6-2 \pmod 6 \\ \mathbf{x} & \equiv & \mathbf{4 \pmod 6} \\ \hline \end{array}\)

The number 4 must go in the center square.

yes, line segment