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Isn't it at Siena?

Where does Laura board the plane, Souly???...I don't see it in the problem..

Not great at this, but  here's  my attempt.....

y = f(2x + 1)  + 3

And we can write

y = 2 ( x + 1/2)  + 3

The "2" horizontally compresses  the  graph  by  a factor of 2

Thus  the  x coordinate of (-2,3)   becomes (-2 / factor , 3)  =  (-2 / 2 , 3)  =  (-1, 3)

The  "1/2"   shifts  the  x coordinate to the left  by  1/2  unit

Thus  (-1,3)  becomes  ( -1 - 1/2 , 3)  =  (-3/2 , 3)

Finally....the "3"  shifts this point up 3 unts   ...so we  have  ( -3/2 , 6)

Thank you!

If the main diagonal is 2......then  we  can find  the side, s   by

2  = sqrt  ( 3s^2)

2 = sqrt (3) s

s =  2/sqrt (3)

s^2  = 4/3

Surface area of this cube  is    6 s^2  =   6 (4/3)  =  24/3  =    8 units^2

And for the sphere radius :

8 =   4 pi r^2

r^2  =  8 / ( 4 pi)  =   2/ pi

r =  sqrt  (2/pi)  units

V (hemisphere )   =  (2/3)pi * (1)^3  =   2pi/3   ft^3

One  of  the smaller hemispheres contains    (2pi/3) / 64 =  (2/192) pi    =  pi/96  ft^3  of chocolate

pi/96  = (2/3) pi * r^3

1/96  = (2/3) r^3

(1/96) (3/2)  = r^3

(3/192)  = r^3

(1/64)  = r^3        take the cube root of both sides

(1/4) ft  =  r 

Looking at Ben's  time  with respect to Philadelphia

2PM  Fri + 14 hrs   =   4AM    Sat  (Philadelphia time)

But Tokyo is  14 hrs  ahead of Philadelphia which  means that  he arrives in Tokyo at 6PM Sat

Looking at Matt's time  with respect to Philadelphia

12 PM  Fri +    12 hr flight + 3 hr layover = 

12PM Fri  +  15 hrs  =    3 AM Sat     (Philadelphia time )

Adding 14 hours to this =  5 PM  Sat   (Tokyo time)

So Matt arrives   first     and  Ben   arrives in  Tokyo   on Sat  at 6PM  

C(6, 3)  = 30 possibilities

3  4  5  6  7  8 

Not all will  work......we must satisfy  the triangle inequality  that says  the sum of any two side lengths of a triangle  must exceed the length of the remaining  side

Possible lengths

3 - 4 - 5    3 - 5 - 6     3 - 6 - 7     3 - 7 - 8     4 - 5 - 6      4 - 6 - 7     4 - 7 - 8     5 - 6 - 7   

3 - 4 - 6    3 - 5 - 7     3 - 6 - 8                      4 - 5 - 7      4 - 6 - 8                       5 - 6 - 8 

                                                                    4 - 5 - 8

5 - 7 - 8       6 - 7 - 8

a)  Probability  is  17/30

b)  There is only 1 right triangle    3 - 4  - 5....probability is  1/30

I'm sorry but this doesn't seem to be the correct answer! Thanks for trying though!

Yep, this one works! In general, you want to be canceling out the \(x^3,x^2\) in order to get a polynomial with degree less than 2. Do you understand why? (note: I'm talking to the original poster)

Answer here :   https://web2.0calc.com/questions/the-point-x-y-is-first-reflected-in-the-x-axis_1

If  we could "unroll"  the cylinder  we  would  have  a rectangle  with a width of 4 ft and a height  of  15  ft

The diagonal  of  this rectangle  represents the  length of the  string  and it is just the  hypotenuse of a right triangle with legs of 4 and  15

Its length  =    sqrt  (4^2  + 15^2 )  =  sqrt  ( 16 + 225)  =  sqrt (441)  =  21 feet

(x + 1)^3  =   x^3  + 3x^2  +3x  +  1

For our polynomial to have a degree less than 2, we  need  to cancel  the  x^3  and 3x^2  terms

So     q (x)   will  do this  if  it is   something like  (-x -3)

Note  that

x^3   + 3x^2 + 3x + 1  +  x^2 ( -x - 3)   =

x^3 + 3x^2  + 3x  + 1  - x^3 - 3x^2  =

3x +  1      and this  is a degree < 2

(note  that other answers are possible ....but this answer for q(x)  seems the easiest )

We need to solve  this

(x + 2)^4  = 500

Note  that    the 4th root  of  500  ≈  4.72   or   -4.72

Ceiling (-4.72)  =  -4

Then we   need  to have    x + 2  = -4

Then  x  = -6   is the lower  bound  integer

And    x = -1   is the  upper bound integer  

So......the   number of points where both coordinates are negative integers  is   6

                                         1  

0.111111111 forever   =   –––  

                                         9  

                                        10  

0.01010101 forever   =   –––  

                                        99

Find a common denominator                    1           11  

and add the two together                        –––   =   –––  

                                                                 9           99  

                     11        10        21                                   7  

                    –––  +  –––  =  –––   which reduces to   –––  

                     99        99        99                                  33  

_.

x^3 ≥  4x

x^3 - 4x ≥  0     (1)

x  ( x^2  - 4)  ≥  0

x  ( x + 2)  ( x - 2) ≥  0

Setting each factor  =  0   and solving for x  produces  three test points

x =  0     x = -2     x   = 2

We have the  intervals

(-inf, 2]       [ -2, 0 ]      [ 0, 2]      [ 2, inf  )

Picking a point in each interval   and testing this in (1)

x = -3    (-3)^3 - 4(-3)   =   -27 + 12  =   -15    and this is not ≥  0....so the first interval doesn't work

x = -1   (-1)^3  - 4(-1)  =  -1 + 4  = 3    and this is ≥ 0....so the second interval works

x = 1    (1)^3  - 4(1) =   -3    and this is not ≥ 0....so  this interval does not work

x = 3    (3)^3  - 4(3)  =  15     and this is ≥  0  ...so this interval works

The solution   is   x =  [-2, 0 ] U  [ 2, inf )

y = (1/5)(3x + 20) →  5y  = 3x + 20  →  3x - 5y + 20  =  0 

We can find the  distance from the center of the circle   (0,0)  to this line with the  following

distance =  l  3(0)  -5(0)  + 20  l               20            

                ___________________  =    ______    (1) 

                sqrt  ( 3^2 + 5^2 )                 sqrt (34) 

A perpendicular line drawn  from the center of the circle to the chord will  bisect  this chord at right angles

 Call the bisection pt,   P     and (1)  =  OP

And 1/2 the chord length  =  10   =  PB

Using the Pythagorean Theorem, we  can find the radius, r, of the  circle  thusly :

r = sqrt  [ (PB)^2  +  (OP)^2  ]

r= sqrt [ 10^2  +  (20/sqrt (34) )^2  ]

r= sqrt  [ 10^2 + (400)/34 ]

r ^2   =   100   + 400/34

r^2  = [3800] 34  =  1900/17

Area of circle  =  pi * r^2  =   (1900/17) pi   units^2

The answer is 7.

The pair (a,b) is (5/2,-8).

Using Phill's diagram...

PR = SR - SP

OR = sqrt(OP² - PR²)

OS = sqrt(OR² + SR²)

y = 2x  means that  the  angle  that this line  makes with the x axis =    arctan (2)

Likewise.....y = 3x  means  that  the  angle   that this line  makes with the  x axis = arctan (3)

To find the slope of  the bisecting line we  can evaluate this  :

tan  [  ( arctan (2)  +  arctan (3) ) /  2 ]  ≈  2.41

See the graph here  :  https://www.desmos.com/calculator/tqi2tckeln  

Considering  the time from  Philadelphias frame of reference

1PM  + 6 hrs  + 1 hr to ceremony location  +  1 hr to get ready  =  9 PM Philadelphis time

But Los Angeles  is  3 hrs  earlier , so  9 PM Philadelphis time = 6 PM LA time......she gets to the ceremony right on  time........!!!!

We are given O = (0,0), A = (5,2) and B = (6, -8). Find [OAB], the area of triangle OAB.

Use Pythagorean theorem to find sides of a triangle OAB

Use Heron's formula to calculate the area of Δ OAB

Find the sum of all values of n that satisfy 1/n-1 + 1/n+1 = 3/n

help is appreciated.

Hello Guest!

\(\color{BrickRed}1/n-1 + 1/n+1 = 3/n\\ \frac{2}{n}=\frac{3}{n}\\ n⇒\pm\infty \)

The function has no zero.

Because point calculation comes before line calculation,

you have to put the divisors in brackets.

\(1/(n-1) + 1/(n+1) = 3/n\)

\(\frac{1}{n-1}+\frac{1}{n+1}=\frac{3}{n}\\ \frac{n^2+n+n^2-n}{n(n^2-1)}=\frac{3(n^2-1)}{n(n^2-1)}\\ 2n^2=3n^2-3\\ n^2=3\\ n=\pm \sqrt{3}\)

\(n\in\{-\sqrt{3},\sqrt{3}\}\)

  !