fiora

(c)I use desmos graphing calculator ,follow the link 

B=2 A=1

Look at hundred place

B+4+A <=14 for 0<=A,B<=5

So, A =0 or 1 

Look at the ones place

Even B=1, the tens place is increment by one 

So let's assume B>=1 ,then we ahve 1+B+1+1=5 \(\Rightarrow\)B=2

Then look at the left parts, 2+4+A=6*A+1\(\Rightarrow\)A=1

Verify it ,it works.

Assume Circle \(\Gamma \) is center at point  \(\Gamma \). Connect \(\Gamma \) to point X and Y

Angle \(\Gamma \)XB = Angle \(\Gamma \)YB =90 degrees.

Angle X\(\Gamma \)Y= 360 - 60 - 90 - 90 =120 degree

And segment \(\Gamma \)X = segment \(\Gamma \)Y \(\Rightarrow \) angle \(\Gamma \)XY =  angel \(\Gamma \)YX =1/2 *(180 degree - angel X\(\Gamma \)Y)=1/2(180-120）=30 degrees.

Angel AYX= angel AY\(\Gamma \) +angle \(\Gamma \)YX=90 degrees + 30 degrees= 120 degrees.

edited: to understand my answer, you need know the properties of inscrided circle(incircle) and angles. I urge you to draw picture to better understand my answer. You should check my answer ,because it might not be correct.

CPhill last answer is right.

The general formula for permutation is \(nPr=\frac{n!}{(n-r)!}\)

In this case \({6}^{P}{5}=\frac{6!}{(6-5)!}=720\).possible combination,that is Jocab has 1/720 chance to open the lock at one time.

A rule of thumb to dealing with permutaion and combination is 

permutation is nPr read as "n pick r", and order does matter.

combination is nCr read as "n choose r", and order does not matter.

See the following online etextbook of probability chapter 3 and 5 for more informations about permutation and combination:

I found the answer with wolframalpha,simply enter these b/d=-0.5-1.5i  , (a+b)/(c+d)=3+4.5i, a/c =1.5-2.5i to wolframalpha to get the answer.However, I still encourage you to give this a try by yourself.

\\S**t, I type my answer then I pressed "post new answer", then all my work gone.

Anyway, denote point (0.62200155,0.78301601) as point B and (0,1) as point A and center point O. connect point A and O,point B and point O.

The length of major arc AB is r*\(\Theta \) where r=1 since it is a unit circle and  \(\Theta \) =asin(y/r)=asin(0.78301601)

r*\(\Theta \)=1*asin(0.78301601)=0.89950000524

you can also verify this with acos=(x/r)

See the following link for visualization:

24 is impossible value of a sector in a unit circle,since the area of unit circle is pi*1^2=pi.

ODEs stand for Ordinary differential equations?You can find most notes about differential equation in the following site.

\(total,amount=\frac{partial-amount}{corresponding-precentage}\)

total amount=1312/(1-20%*30%-80%*15%)=1312/82%=1600

First,nice questions,MaxWong.

For (1) I want to add some step from the equation given by the guest\(log(b)=-({x}^{2}*log(a))/(x+1)=-({x}^{2}/(x+1)*log(a))=log({a}^{-\frac{{x}^{2}}{(x+1)}})\)

Now,cancel logarithms by taking exp with both sides:

\(b={a}^{-\frac{{x}^{2}}{(x+1)}}\)

(2)I assume you mean \(a=log{2}^{3} , b=log{3}^{7}\)instead of x and y

\(log3=a*log2,log2=\frac{log3}{a}, log7=b*log3\)

\(log{42}^{56}=log{42}^{42*4/3}=1+log{42}^{4/3}=1+log(4/3)/log42=1+\frac{log(4/3)}{(log2+log3+log7)}\)

Therefore,\(log{42}^{56}=1+\frac{log(4/3)}{(log3)/a+a*log2+b*log3}\)

(3), Given that a>1 and b>1 and \({a}^{x}*{b}^{y}={a}^{y}*{b}^{x}=1\), prove that x + y = 0.

\({a}^{x}*{b}^{y}=1\Rightarrow log({a}^{x}*{b}^{y})=log1=0\)

\(x*log(a)+y*log(b)=0 \)(1)

Same for the second equation,we have

\(y*log(a)+x*log(b)=0\)(2)

combine (1) and (2),we have

\((x+y)(log(a)+log(b))=0\)(3)

Since a>1 and b>1,so log(a)+log(b)>0

So,the equation (3) is true if and only if x+y equal to 0

Hence,x+y=0 if a>1 and b>1