All Answers 358467 Answers

Melody made a small typo. =0.18g = 1.8 x 10^-1g.

1- Expand the following:
3^(n - 1) (n - 1) = n×3^n

Express 3^(n - 1) (n - 1) as a difference of fractions.
3^(n - 1) (n - 1) = 3^(n - 1) n - 3^(n - 1):
3^(n - 1) n - 3^(n - 1) = n*3^n

2- Expand the following:
-3^(n - 1) (2 n - 7) = 3^n (5 - 2 n)

Express -3^(n - 1) (2 n - 7) as a difference of fractions.
-3^(n - 1) (2 n - 7) = -(3^(n - 1) (-7)) - 3^(n - 1)×2 n:
-(-7×3^(n - 1)) - 2×3^(n - 1) n = 3^n (5 - 2 n)

Multiply -1 and -7 together.
×3^(n - 1) - 2×3^(n - 1) n = 3^n (5 - 2 n)

Distribute 3^n over 5 - 2 n.
3^n (5 - 2 n) = 3^n×5 + 3^n (-2 n):

7×3^(n - 1) - 2×3^(n - 1) n = 5×3^n - 2×3^n n =(5 - 2n)*3^n

Dont be so lazy.

Type your question in, it is only words. 

I get these to be:

s_n = n3ⁿ   and t_n = (5 - 2n)3ⁿ  where n starts from 0.  (i.e. n = 0, 1, 2, 3, ...)

Here are the 4 extra acute angles

There is a point inside a polygon such that if you rotate the polygon around that point clockwise by 55 degrees, the polygon coincides with itself. What is the minimum number of vertices that this polygon can have?

I'm going to guess a bit.

I think the polygon probably has to be regular (all sides and angles the same)  And the rotation point is in the middle. 

360/55 =72/11 = 6.5454545454545455 sides but it has to have a whole number of sides so this is no good

72/11*11 = 72

Well if it has 72 sides this will work.

360/72 = 5  So every 5 degrees it will rotate onto itself.   So it will overlap itself 11 times in a rotation of 55 degrees.

Anyway, my guess is 72 sides. Can someone do it with less sides?

Thanks Chris!

In general:  \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)  so we can also do it like this:

\((2x-\frac{3}{4}y^2)^3\rightarrow (2x)^3-3(2x)^2(\frac{3}{4}y^2)+3(2x)(\frac{3}{4}y^2)^2-(\frac{3}{4}y^2)^3\\ \rightarrow 8x^3-9x^2y^2+\frac{27}{8}xy^4-\frac{27}{64}y^6\)

where \(a=2x \text{ and } b=\frac{3}{4}y^2\)

.

!!!!!!!!!

Neat problem.

First off note that \(\dbinom{n}{k}=\dbinom{n}{n-k}\)

We are choosing (n-1) items from 2n possible items.  One way to do this is to split the 2n items into 2 piles of n items each, choose k items from the first pile, and (n-1-k) items from the second pile.  We do this for \(0 \leq k \leq n-1\)

This gets us

\(\dbinom{2n}{n-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{n-k-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{k+1}\)

you can recognize the right hand term as the summation shown

The weight of a honeybee is 1.2 x 10-1g. The weight of the pollen collected by the bee on one trip is 6.0 x 10-2g.  What is the combined weight of the bee and the pollen?

 1.2 x 10^-1 = 0.12

 6.0 x 10^-2 = 0.06   (edited)

add             = 0.18g     or 1.8*10^(-1) g      (edited)

1)-

 S(0) =0
S(1)=3
S(2)=18
S(3)=81
S(4)=324
S(5)=1215
S(6)=4374
S(7)=15309
0, 3, 18, 81, 324, 1215, 4374, 15309, 52488, 177147, 590490, 1948617, 6377292, 20726199, ...etc.
a_n = 3^(n - 1) (n - 1)  - The closed form.

2)-

t(0)=5
t(1)=9
t(2)=9
t(3)=-27
t(4)=-243
t(5)=-1215
t(6)=-5103
t(7)=-19683
5, 9, 9, -27, -243, -1215, -5103, -19683, -72171, -255879, -885735, -3011499....etc.
a_n = -3^(n - 1) (2 n - 7) - The closed form.

What is the degree measure of the smaller angle between the hour hand and the minute hand of a clock at exactly 2:30 p.m. on a 12-hour analog clock?

At 2:30 the big hand is on the 6 hour mark and the little hand is half way between the 2 and the 3

The clock is divided into 12 hourly segments so each segment is 360/12 = 30 degrees

the hands are 3.5 segments apart

3.5*30 =  105 degrees

Let the time taken by Terry=T
Time taken by Tao =T - 5
60T + 40(T - 5) =2,000
60T + 40T - 200=2,000
100T =2,200
T = 22 minutes after 2 pm when they will meet.
Note: EP made a typo of 23 minutes.

Terry gets a 5 minute head start and covers 5 min x 60 m/min = 300 m

  then Tao starts at 1405       They have  2000 - 300 = 1700 meters to cover before they meet

     the distance is closing at (40+60)= 100 m/min

         1700 m  / (100 m/min) = 17 minutes after Tao starts   14:05 + 17 = 14:22    (02:22 PM)

                                                                                      (corrected...THANX!)

Hour hand moves 360 degrees in 12 hours  ....   30 degrees per hour or 15 degrees per 1/2 hour

   so 2:30  would be 5    1/2 hour segments   5 x 15 = 75 degrees= angle of hour hand.    The MINUTE hand will be at 180 degrees (pointing directly downwawrd)      180 degrees - 75 degrees = 105 degrees.

the above answer sort of misses the point I think.

We can notice that \(k^3 \pmod{6} = k,~k=1,2,\dots 5 \\ \\ \text{and obviously }6^3 \pmod{6} = 0\)

so we have a repeating sequence in the sum \( 1+2+3+4+5+0 + 1+2+\dots\)

there are 16 of these groups of 6 and then 4 additional elements of the sum

so we have

\(\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 3\cdot 16 + 1+2+3+4 = 58 \\ \text{and} \\ \\ 58 \pmod 6 = 4 \\ \\ \text{so }\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 4\)

2)-

All numbers from 27 to 35 in base 10 are:
{1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1022}in base 3
{43, 44, 45, 50, 51, 52, 53, 54, 55}in base 6
And the average is 31 in base 10

1)  333^(5n - 2) mod 11 = 5 always, since:

(5n -2) =333, when n = 67, it follows that:

333^333 mod 11 =5