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Thanks!

See answer here:

https://web2.0calc.com/questions/the-numbers-from-1-to-11-are-drawn-at-random-find

1st post she walks 2m to put the post in, then 2m back to the pile = 4m

2nd post is at the pile = 0

3rd  = 4

4th  = 8

5th =  12

nth one = (n-2) *4

Total distance =  4 + \(\sum_{2}^{200}\)(n-2)*4  meters     (this assumes she walks back to the pile location after the final post placement)

78,808 m  =  78.808 km

IF she doesn't walk back to the pile it is 396 m LESS.

CPhill i think theres some problem in the 'c' parts answer becoz my teachers answer is 144.

its somewhat like:

No. of ways in which none of U , A , E are together = 3! x ⁴P₃  = 3! x 4!/1! = 3! x 4! = 144.

i have no idea what is done here.

dont you think that U, A , E ..that are vowels...and when they wont appear together, then the answer will be the same as the 'b' part???? 

thanks hectictar for noticing..i appreciate it!thank you! 

Hey rosala, your link doesn't need the  " /new "  at the end...it should just be

https://web2.0calc.com/questions/permutations-and-combinations-lt-3

Just wanted to prove GA's answer in an algebraic manner

r = rate of current  (in m/s) =  rate of hat

R  = Andy's normal rate  (in m/s)

Distance from the bridge  hat has traveled in 5 min since being knocked off  =  300 (r)  =   300r  m

Distance from bridge Andy travels in 5min = 300(R - r) m

So....when Andy turns around ...his new rate is  R + r

And the distance that Andy travels to retrieve the hat after he turns around  =  [300(R - r) + 1000] m

Meanwhile.....while Andy travels this distance, the hat  travels  [1000 - 300r ] m

So....equating times, we have that

Distance Andy travels after turning around  / His Rate  =   

Distance hat has traveled / hat's rate

So we have that

[ 300(R - r) + 1000 ]  / [R + r]   = [ 1000 - 300r ] / r

[ 300R - 300r + 1000] r   =  [ 1000 - 300r ] [ R + r ]

300Rr - 300r^2 + 1000r  =  1000R - 300Rr  + 1000r  - 300r^2

600Rr  - 1000R   = 0

3Rr  - 5R  = 0

R ( 3r - 5)  = 0   ⇒   3r  = 5  ⇒    r  =  5/3   =  1.67 m/s

Just  as GA found  !!

We have 5 even numbers

We can choose any 5 of the even numbers first.....

And then any 4

And then any 3

And then any 2

And then the last

So....the probability is

(5/11) (4/10) (3/9) ( 2/8) (1/7)  =

(5/11) (2/5) (1/3)(1/4)(1/7)  =

10 / 4620    =   1 / 462

Another way of looking at this is to note that we have 5! ways of arranging the even digits first and 6! ways of arranging the odd digits second

And the total possible arrangements =  11!

So....the probability is  

5! * 6!  / 11!    =    1 / 462

The triangles don't have the same height.....they are bases of prisms that have the same height

We have that :

(x + 3)  + d  =  (3x - 10)       (1)

(3x - 10) + d  = (2x + 10)     (2)

Subtract  (2)  from  (1)   and we have

( x + 3)  - (3x - 10)   = (3x - 10) - (2x + 10)     simplify

-2x + 13   =  x - 20      add 2x, 20 to both sides

33   =  3x       divide both sides by 3

11  = x

We need to find the radius of the circle....so we have that

8 pi    =  2pi * r       divide both sides by 2pi

4 cm  = r

(1/3)  of the area of this triangle can be found  as  :

(1/2) r^2 sin (120° )  =

(1/2) 4^2 * √3/2  =

4√3  cm^2

So....3 times this is  just

12√3  cm^2  ≈  20.78 cm^2

Once Andy stops rowing, his relative velocity and direction are the same as the hat. Rowing away from the hat for five minutes means it took five minutes for him to return to it, for a total of ten minutes. In that ten-minute length of time, the hat traveled one kilometer. 

Current flow: 1 km/10 min = 100 meters per minute

= 1.67 meters per second.

GA

 f(x)   =   2x + 5

g(x)   =   √[ f(x) ] - 2

h(x)   =   f(g(x))

h(2)   =   f(g(2))

g(2)   =   √[ f(2) ] - 2

 f(2)   =   2(2) + 5   =   4 + 5   =   9

g(2)   =   √[ f(2) ] - 2   =   √[ 9 ] - 2   =   3 - 2   =   1

h(2)   =   f(g(2))   =   f( 1 )

 f(1)   =   2(1) + 5   =   2 + 5   =   7

h(2)   =   f(g(2))   =   f( 1 )   =   7

t(-4) = |-3+2(-4)|  =| -3-8| = 11

t(11) = |-3+ 2(11) | =19

19

I'm still somewhat confused. Sorry. The question askes for both triangles to have the same height so wouldn't be 40/40=36/30 or something like that and then find the volume somehow.

Sorry Im really bad a geometry and I really appreicate you taking the time to help me out. :) 

When the horse gets to a corner of the fence that is adjacent to the corner where the rope is attached, he will have  8m  of rope left. So there are two quarters of a circle with a radius of  8m in the grazing area. And there are three quarters of a circle with a radius of 18m in the grazing area.

grazing area   =   (3/4)(pi * 18²)m²   +   2(1/4)(pi * 8²)m²  

grazing area   =   (3/4)(324pi)m²   +   2(1/4)(64pi)m²

grazing area   =   243pi m²   +   32pi m²

grazing area   =   275pi m²

Here is a link to a picture because it didn't show up well in the post:

(x+3)+ (3x-10) + (2x+10) =6x + 3

Sum them up as an arithmetic series:

6x + 3=3/2 *[2*[x+3] + (3-1)*(2x - 13)], solve for x

Solve for x:

6 x + 3 = (3 (2 (x + 3) + 2 (2 x - 13)))/2

6 x + 3 = (3 (2 x + 6 + 2 (2 x - 13)))/2

6 x + 3 = (3 (4 x - 26 + 2 x + 6))/2

Grouping like terms, 4 x + 2 x - 26 + 6 = (2 x + 4 x) + (6 - 26):

6 x + 3 = (3 ((2 x + 4 x) + (6 - 26)))/2

6 x + 3 = (3 (6 x + (6 - 26)))/2

6 x + 3 = (3 (6 x + -20))/2

Multiply both sides by 2:

2 (6 x + 3) = (2×3 (6 x - 20))/2

(2×3 (6 x - 20))/2 = 2/2×3 (6 x - 20) = 3 (6 x - 20):

2 (6 x + 3) = 3 (6 x - 20)

Expand out terms of the left hand side:

12 x + 6 = 3 (6 x - 20)

Expand out terms of the right hand side:

12 x + 6 = 18 x - 60

12x - 18x = -60 - 6

- 6x = - 66

x = -66 / -6

x= 11

The number is a very large prime find how many digits it would take to write the number in decimal form. ( Hint: Log 2 is an interesting irrational number It's the power of 10 that gives 2, in the same way, that 3 is the power of 10 that gives 100. Your calculator can calculate log 2)

\(let \\\;\;y=2^{3217}  \\ log(y)=log(2^{3217} )\\ log(y)=3217\;log(2 )\\ log(y)\approx3217*0.301029995\\ log(y)\approx\;968.4134961\\ 10^{log(y)}\approx\;10^{968.4134961}\\ 10^{log(y)}\approx\;10^{0.4134961}*10^{968}\\ y\approx\;2.59*10^{968}\\ 2^{3217}\approx\;2.59*10^{968}\\ 2^{3217}-1\approx\;2.59*10^{968}\\\)

This is an integer, so the number is 2 followed by 968 more digits.

So the the number is 969 digits long

Congratulations penquino  !!    

1) a.  In acute triangle ABC, we know AB = 7, BC = 8, and that Line{CA} is the shortest side. What is the smallest possible integer value of CA?

The angle opposite  the longest side must be < 90°

Therefore

√[ CA^2  + 7^2 ] > 8^2

CA^2  + 49  > 64

CA^2  > 15

CA > 3.8

So....the smallest possible integer value of CA  is  4

b.  In obtuse triangle ABC, we know AB = 7, BC = 8, and that Line{CA} is the longest side. What is the smallest possible integer value of CA?

We must  have  that

AB^2  + BC^2  <  CA^2

7^2  +  8^2  <  CA^2

49  +  64  <  CA^2

113 < CA^2        take the square root of both sides

10.63  < CA   ⇒   CA  > 10.3

So....the shortest  integer side length for CA  is   11

2)  a.Two diagonals of a parallelogram have lengths 6 and 8. What is the largest possible length of the shortest side of the parallelogram?

The diagonals will bisect each other.....so we have a triangle with sides of  3 and 4

The largest possible (integer) length of the shortest side, S, of the parallellogram  must be

3 + S  >  4

S  >  1

So...the largest possible (integer) length of the shortest side  is  2

b.  Two sides of an acute triangle are 8 and 15. How many possible lengths are there for the third side if it is an integer?

Let S be the missing side...so we have....

S + 8 >  15

S >  7  so the shortest integer length is 8

But also...since it's an acute triangle,  the longest possible side is

√ [ 8^2 + 15^2 ]  > S

17 >  S  ⇒   S <  17

So....the greatest possible integer  length of S  is  16

So...the number of possible lengths  is   16  - 8 +  1      =  9

Sum them up as an arithmetic series:

N =[L - F] / D + 1

10 =[L - 12] / 5 + 1

L = 57 - last allowance

[F + L] / 2 x N =Sum

Sum =[12 + 57] / 2 x 10

Sum =[69] / 2 x 10

Sum = 34.5 x 10

Sum = $345.00 - Andrew's total allowance over 10-week period.

1.  12 girls - 8 boys  =

We can choose   any 12 of the 20  to be girls  ....so we have ...

C(20, 12) * .5^20   ≈ .12   ≈  12%

2.  6 boys - 14 girls

We can choose  any 14 of the 20 to be girls...so we have

C(20, 14) * .5^20  ≈  .0369  ≈   3.7%

-2/3 is the answer.

( x , y )   reflected across the x axis produces  ( x , - y)

So

(-3 , 7 )  ⇒  ( - 3 , - 7)

1. y  =  x^2 - 4x + 1

2.  y  = .782x^2 - 20.213x + 179

3.  y =  0.032x^2 - 0.601x - 0.255

4. y  =  -3.5(3)^2  - 0.5 (3) + 65  =  32  in

5. y  = 1.65 (8)^2  + 18.25(8)  + 155   =  406.6  =  $407, 000