fiora

AD=AE\(\Rightarrow\) triangle ADE is an isosceles triangle. \(\Rightarrow\) angle ADE= angle AED=(180-56)/2=62 degrees

BA || CE \(\Rightarrow\) angle BAD = angle ADE = 62 degrees

BA || CE \(\Rightarrow\) BA || CD and CB || DA by given \(\Rightarrow\)quadrilateral BCDA is a parallelgram

angle BCD = angle BAD = 62 degrees (opposite angles of parallelgram are equals) 

assume all female have math degree.1300 out 2000 people have math degree, while 640 of those 2000 people are female. Therefore,the minimum number of males with maths degrees is 1300-640=660

 If the last three digits of a whole number are divisible by 8, then the entire number is divisible by 8. (By googling)

I don't have an effective way to determine a 3 digit number wehether is multiple of 8 or not.

So I list all 3 digit number consist only numbers 1-6,which are

112,136,144,152,216,232,224,256,264,336,344,312,352,416,456,424,464,432,536,544,512,552,616,624,632,664,656

27 numbers divisible 8 in total  (I did that by adding multiple of 40, so it is  a bit out of order.)

I also wrote a stupid Java program to check my answer.

public static void main(String []args){
        int c=0;
        for (int i = 1;i<=6;i++){
            for (int j = 1;j<=6;j++)
            {
                for(int k = 1;k<=6;k++)
                {
                    int n=i*100+j*10+k;
                    if(n%8==0)
                    {
                        c++;
                    }
                }
            }
        }
        System.out.println(c);
     }

The first five dight can be any integer in {1,2,3,4,5,6}.Therefore, the probability that the number formed is a multiple of eight is \(\frac{6^5*27}{6^8}=\frac{3^3}{6^3}=\frac{1}{2^3}=\frac{1}{8}\).

PS: I believe there is a way to do this problem in abstract algebra, which I rusted at.

Assume the quadrilateral is a square, then AB = BC = CD = DA =a for some non-negative a

AM = BM = BE =EC = CF = FD = 1/2*a (given)

Triangle MAD is congruent to triangle EBA also congruent to FCB (Side-angle-side)

Therefore, angle EAB = angle MDA, angle AMD = angle BEA (corresponding angles are equal)

angle EAB + angle BEA = 90 degrees \(\Rightarrow\)angle MDA + angle BEA = 90 degrees \(\Rightarrow\) angle AND and angle MNA are right angle

That is triangle MNA is similar to triangle AND and triangle MAD

MA / AD =1/2  \(\Rightarrow\) ND = 2*AN and AN = 2* MN \(\Rightarrow\) DN / MN =4

Similarly, you can prove angle BGE is a right angle, which yield Trinagle BGE is similar to BCF (Angle-Angle)

BF=\(\frac{\sqrt{5}}{2}a\)  by pythagorean theorem, cos of angle CBF = BC / BF = \(\frac{2}{\sqrt{5}}\), BG = BE *cos CBF =1/2 * a*2/sqrt(5)=\(\frac{\sqrt5}{5}a\) 

GF= BF - GF =\(\frac{3\sqrt5}{10}a\), GE=1/2*BG=\(\frac{\sqrt5}{10}a\),FG/GE = 3 , Triangle BGE is congruent to triangle ANM, which yield MN =EG

Therefore FG/MN= FG/GE = 3

Connect BN, AG = AB*cos BAG = AB * cos CBF = \(\frac{2}{\sqrt{5}}a\)

Triangle BAG is similar MAN. AM/AB=1/2 \(\Rightarrow\) AN= 1/2 AG \(\Rightarrow\) GN= 1/2*AG = \(\frac{\sqrt5}{5}a\)

BN= \(\sqrt{{​​(\frac{\sqrt5}{5}a)}^{2}+{​​(\frac{\sqrt5}{5}a)}^{2}}=​​\frac{\sqrt{10}}{5}a\) by  pythagorean theorem, and MN =EG=\(\frac{\sqrt5}{10}a\)

BN/MN= \(2\sqrt{2}\)

Edit: highlighted answer

Simply serach it on the internet.  LOL

Because this question show up on my complex analysis test, and the test covered the topic of complex sequence and series.But why is this on my test? This is also my question for my wired professor. Complex number include real number(complex numberz=x+i*y, for real x and y, if y=0 then z is real number), so the definition can also apply to (2). The definition is indeed in my complex analysis textbook, and normally I can use this defintion to proof the sequence converge or diverge. But in this case, I guess I can show the sequence is bounded and then by some theorem/corrollay to show thatthe sequence is converge. If don't know solve this with any kind of math, just leave this alone and don't bother the defintion stated in my original question. Talk is cheap, show me the math(code).--Linus Trovalds 

cos(u-v)=cos(u)cos(v)+sin(u)sin(v)

given sin(u)=12/13 , cos(v)=-4/5

The given inforamtion is not enough to find the unique answer.

Assume cos(u)=5/13,sin(v)=3/5

cos(u-v)=-4/5*5/13+3/5*12/13=16/65

Assume cos(u)=5/13,sin(v)=-3/5

cos(u-v)=-4/5*5/13+(-3)/5*12/13=-56/65

assmue cos(u)=-5/13,sin(v)=3/5

cos(u-v)=-4/5*(-5)/13+12/13*3/5=56/65

assume cos(u)=-5/13, sin(v)=-3/5

cos(u-v)=-4/5*(-5)/13+12/13*(-3)/5=-16/65

You can find more information in the following link, for this topic read chapter 16

Thanks for verification,guest. My professor like to give the student some wired question.He is differently something else and very infamous in my University. He even taught the co-chair of Department of Math & Computer Sicence of my University,so you can imagine his age. I am pretty sure that sample problem in the link is the best and the only sample I can found.Thanks again for your suggestion. 

Thanks guest fort spoting my mistake. I am sorry, I was doing another homework and didn't double check my typing.Yeah , you are right，the correct formula for sin(z) is sin(z)=sin(x+iy)= sin(x)*cosh(y)+i*cos(x)sinh(y). The image of strip R under mapping f(z)=sin(z) is  quadrant+the fourth quadrant+the positive real axis of complex plane. Let w=f(z),then Re(w) is the set positive number and Im(w) is real number.Then the image is an open set,isn't it? Where the definition of open set is  A set is open if all its points are interior points,and the definition of interior point is A point a ∈ G is an interior point of G if some open disk with center a is a subset of G. And open disk is the inside of this circle with center a and radius r; we use the notation D[a,r] := {z ∈ C : |z − a| < r} .

See a similar problem posted on stack exchange

Thanks,Rom and Guest. Hey Rom,I think your answer is partially correct.

The mapping f(z)=sin(z)=sin(x+iy)=sin(x)cosh(y)+icos(x)cosh(y)  map z to Real(f(z)) in positive real number and Im(f(z)) to real, which map to first and fourth quaduant and real axis of complex plane. That is my simpilfie answer and this probelm was one of test question,and I am doing test correction. Does anyone have any thoughts on this?   

135 degree

I can't explain due time being.

2000-1000=1000

(a)10700

(b)10875

(c)11100

(d) I think Mrs. Solar should choose contractor should choose contractor 3. beacuse contractor finish the work faster.

Mrs. Solar can eliminate contractor 1,beacuese Contractor 1's time estimate is 210 h which larger than Mrs. Solar expected.

However,the appropriate answer is Mrs. Solar should choose contractor 2 beacuse contractor 2's rate of charge 10875/195 is less than contractor 3's rate of charge 60.

Hey, you have not given the charge rate per hour of contract 1 and 2. Unable to solve due to incomplete information.

Sorry I can't edit my previoud answer and it was incomplete. You should extend the line to the end and put an arrow at the front section.

That shuld be fine, but I think you need to extend the line to the end put an arrow at the front section of the line,since 0