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So I actually thought many times about the answer now and I got a question myself.

The answer under me seems mathmatic right, but it still wondered why the wind would hinder the plane in a ration of 12/18. ( As its 12minutes faster to the target ,and 18minutes late to get back to it)

If I wouldnt think about the speed of the plane in d/h, but in time I could create the following equations:

W = a negative number which shows how many minutes the winds let u reach ur destination faster

A = the time the Plane needs to reach its destination. always the same as stated in the topic

So with A being always the same + having the advantage of the wind it takes u 60mintues.

A+W = 60minutes

Now with A and the disadvantage its 90 minutes.

A-W  = 90minutes

Means if i solve this equations:

A+w =60minutes

A-w =90minutes

2w = -30minutes      II /2

w  = -15minutes

putting this in the first equation

so A-15minutes = 60minutes

shows that A is 75minutes.

In the answer under me though its showed that A is 72minutes, when solved with involving the destination.

Im really confused here maybe someone can help

Thanks so much for your help! I appreciate it lots. :) How can I post a picture of problems (1) and (3)?

Just want to point out that my answer is incorrect.

I really didnt read the topic correct, ones under me is correct thou.

There is no largest prime number. The primes go on forever!

1,000 / (1/5) =1,000 x 5 =5,000.

The largest prime known to date is: 2^74,207,281-1 with 22,338,618 digits long.

idk 

i agree with CPhill, but the question says round to the nearest whole number, so it would be 12%

2hydrogen atoms : 1 oxygen atom   (h2o)

3*10^23 atoms     II /3 because every 3rd atom is oxygen

means

1*10^23 atoms are oxygen

So the rest of

2*10^23 atoms are hydrogen.

Which means u got 10^23 H2 Atoms.

Let the airplane's speed = A

Let the wind's speed =W

Let the distance between Sydney and Melbourne =D

A + W =D/1 hour

A - W =D/1.5 hours, solve for A

A =5/6D hours - That is, the airplane absolute speed regardless of wind, is 5/6 the distance between Sydney and Melbourne. So the scheduled time for the trip must be:

D / (5/6D) = 6/5 hours

D =6/5 hours, or 72 minutes. But the returning plane took 90 minutes, therefore:

90 - 72 = 18 minutes too late. 

7:30 start 1

8:30 arrival 1

9:15 start 2

10:45 arrival 2

I dont really get the question. Kinda sounds like a trick question, but if ur just asking for the time it was late in the 2nd arrival, wouldnt that just be

(2 time - arrival 2 time) - (start1 - arrival1) ?

1h30m - 1h = 30 minutes 

Of cours by saying that the airplane was on time in the first flight, and not too early or also too late.

What is 48/2(9+3)?

If there is a fractional element (number, variable, parenthesis expression) without a multi-signum, they are a product and are continued as a product.
These elements can not be simultaneously in the numerator and in the denominator.

2(9+3) is clearly a product.

That's why 

 \(48/2(9+3)=\frac{48}{2(9+3)}=\frac{48}{24}=2\)

  !

Volume of pitcher  =  [40 cm x 40 cm  x 26 cm ]  = 41600cm^3

1L   = 1000cm^3

The pitcher easily holds 1L

9/16 because you need to multiply the number; 3/4; two times

Merry Christmas to you too! 

Merry Christmas......Piperson  .......!!!!!

( 1)  and (3)  can't be answered without a figure  shown.....!!!!

(2)    9 * (3/2)  =  27/2   =   13.5 units long after  dilation

4. Point J is located at (2,5)  and point K is located at (4,19). What are the coordinates of the point that partitions the directed line segment JK in a 3:2 ratio?

The line is divided into 5 parts........the partition point will be located at  3/5 of the way between (2,5) and ( 4, 19)  =   [ 2 + (4 − 2)(3/5) ,  5 + (19 − 5) (3/5)  ]   =  [ 2 + 6/5 , 5 + 42/5]   =  [ 16/5 , 67/5]

Proof.....distance beween  ( 2,5)  and (4, 19)  = √[ (4 − 2)^2 + ( 19  − 5)^2 ]  =  √ [ 4 + 196]  = √200 = 10√2

Distance beween  ( 2,5)  and (16/5, 67,5)  = √[ (16/5 − 2)^2 + ( 67/5  − 5)^2 ]  =

√[ (6/5)^2 + ( 42/5 )^2 ]   =  √ [ 36  + 1764] / 5  =  √1800 / 5   = 30√2 / 5 =  6√2

And    6√2  /10 √2  =   6 / 10   =  3/5

Volume = 4/3 x Pi x r^3, r=radius

3 sig figs

Here are the rules :  

http://people.uwec.edu/carneymj/Adobe%20handouts/Significant%20Figures.pdf

z - 1/2 = 1/10     add 1/2 to both sides

z = 1/10  + 1/2           [1/2 =  5/10]

z = 1/10 +  5/10  

z  = 6/10   =   3/5

Same to you.......penquino  !!

Hi Max,you realise that you are dealing with the product rule and chain rule here,but d/dx sin(1/x) 

=  { sin(1/x+h) - sin(1/x) }   /h.    Now use your formula sinA - sinB -2cos(A+B)/2 .sin(A-B)/2 to get

2cos{(x+1/2h)/x(x+1/2h) } .sin {(-1/2h)/ x(x+1/2h)

-----------------------------------------------------------------

               h               lim h tends to zero

divide numerator and denominator by 2    to get

cos{ (x+1/2h/x(x+1/2h) .sin{ (-1/2h)/x(x+1/2h)

-----------------------------------------------------------             since sin (x)/x tends to 1 in lim h tends to zero

         1/2 h

you are left with cos  { (x+ 1/2h)/x(x+1/2h)  which reduces to cos (x/x^2) as h tends to zero which gives cos(1/x).

You can go on to prove the product and chain rules to validate the rest of the derivative.Hope this helps.