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The distance from a point ( m, n)   to  line in the form  Ax + By + C  = 0  is given by :

l A(m) + B(n)  + C l

________________

√ [ A^2 + B^2]

So....  A = 2,   B = 1  and C = - 5

The formula for the distance from a point to a line is \(d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}\) when a line is written in the form \(Ax+By+C=0.\)

Cphill used the given equation for the line \(2x+y-5=0\). Therefore, \(A=2, B=1, \text{ and }C=-5\)

thank you very much! may i ask where u got √[ 2^2 + 1^2 ] from?

Thanks, tertre!!!!

That IS easier to use.....!!!

We have these points on the graph   (0,0)  and ( 3, - 2)

The slope  is   [ -2 - 0 ] / [ 3 - 0 ]   =   -2/3

So.....using the slope and (0,0).....we have the equation

y = (-2/3) ( x - 0) + 0

y = - (2/3)x        multiply through by 3

3y  = -2x      add 2x  to both sides

2x + 3y   = 0

5x + 8y = -9

In the form Ax  + By  = C

The slope is  -A / B

So....the slope of L1  is   -5/8

And L2  is a perpendicular line to L1.....so.....it has a negative reciprocal slope  =  8/5

So.....the equation of L2   is

y =  (8/5) x - 10) + 10

y = (8/5)x - 16 + 10

y = (8/5)x  - 6

So

m + b   =   (8/5)  - 6   =   (8/5) - 30/5    =    - 22/5

Let the time taken by Co. B = B, then:
The time taken by Co. A = B+2
1/B + 1/(B+2) =1/24, solve for B
Solve for B:
1/B + 1/(B + 2) = 1/24

Bring 1/B + 1/(B + 2) together using the common denominator B (B + 2):
(2 (B + 1))/(B (B + 2)) = 1/24

Cross multiply:
48 (B + 1) = B (B + 2)

Expand out terms of the left hand side:
48 B + 48 = B (B + 2)

Expand out terms of the right hand side:
48 B + 48 = B^2 + 2 B

Subtract B^2 + 2 B from both sides:
-B^2 + 46 B + 48 = 0

Multiply both sides by -1:
B^2 - 46 B - 48 = 0

By using the Quaratic Formula, we have:

B = 23 + sqrt(577)- Hours      and A = 25 + sqrt(577)- Hours.

2x + y - 5 = 0

(x-3)^2 + (y - p)^2  = 5

The center of the circle is  (3, p)

And the distance from this center to the point of tangency  is  √5  units

Using  the equation for the distance from a point to a line, we have

l  2(3) + 1(p) - 5 l

______________   =    √5

√[ 2^2 + 1^2 ]

l 6 + p - 5 l      =   √5

_________                     multiply both sides by √5

√5

l  6 + p  -  5 l   = 5

l 1 + p l   = 5

And we have two equations

1 + p  =  5             and        1 +  p   =   - 5

p = 4                                     p = -6

So.....the  coordinates are   (3, 4)    or  (3, -6)

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

Deleted.

To answer the first problem, we can use the clock formula, which is \(|30H-5.5M|\). Therefore, plugging the values in, we yield with \(|30(9)-269.5|=270-269.5=\boxed{.5}\)

You can't. Your message is set in stone. It is eternal, undying, engraved in the code of this site. Your message will not go anywhere, it will stay here for all eternity, until the end of time itself.

You have made a mistake, ANSP, an unforgiveable sin, and now you ask for our forgiveness? now you ask to fix what you have done? You have doomed us ANSP, you have doomed us all, and for your deadly sin, we are all going to PERISH.

Yes I need the answer to this too gueesstt

Thanks, GA!!!

But.....I have to admit......you wouldn't have confused me any more than if you had provided  detailed instructions as to the assembly of the space shuttle.....LOL!!!

Do you have a specific problem????

Yep......you just had a slight case "brain - freeze "   .....LOL!!!

Thank you!!

Easiest way to do this is to factor top/bottom  separately

9 - x^2  =  (3 - x) ( 3 + x)  

9x + 27  =  9 ( x + 3)

So we have

(3 - x)  ( 3 + x)                     (3 - x) ( x + 3)                (3 - x)

____________        =       _____________   =       ________

9 ( x + 3)                             9 (x + 3)                            9

Thank you! That was simplier than I first thought the problem would be. The two x-4s just cancel out.

1/3 - x/9   is what I find.....but I do not see that option....

4mn^6                  4        m          n^8

________  =      ___  x  ___  x   _____   = 

28m^5n^3           28       m^5      n^3

1    x     1     x    n^5

__       ___      ____  =

7         m^4        1

  n^5

______         [  Do you understand what I did, Phoenix  ???  ]

7m^4

Thanks CPhill! Yes, let's hope the weather does not get much colder!!! At least you get a break from Louisiana sun. It's going to be a freezing Oregon winter here lol.

Just "cancel"   the  " x - 4"  factor on top/bottom  and we end up with

x + 2

_____

x + 4

We have a "rational" function here, Phoenix  [ a ratio of functions ]  

The domain restrictions  are the values of h that makes the denominator = 0

So

h^2 - 12h + 20   =   0      factor

(h - 10) ( h - 2)  = 0

Set each factor to 0

h - 10  = 0                    h - 2  = 0

h = 10                          h = 2

So....h = 2 ,  h =10   are the domain restrictions

HAHAHA!!!!!. I live in Louisiana.....after our brutal Summers, a bit of cold weather is welcome....but......not too cold!!!

The DENOMINATOR cannot equal ZERO

The denominator factors to   (h-2)(h-10)       so h   (or x) cannot equal 2 or 10....all other values of h (or x) are allowed.

At 9 o'clock......hands are separated by 1/4 * 360  =  90°

And, at this time.....let the minute hand be at 0°

And let the hour hand be at 270°

In 49 minutes.....the hour hand moves 360/12 * 49/60   = 24.5°

So.....the hour hand is at   [270 + 24.5]   =  294.5°

And the minute hand has moved  [ 0 +  49/ 60 * 360 ]  = 294°

So....they are separated by 0.5°  =  (1/2) of a degree

To find the exact time when they are together....

The hour hand moves at  30/60  =  .5 degrees per miniute = (1/2)° per minute

The minute hand moves at (360/60) = 6° per minute

So....we need to solve this

294.5 + (1/2)M  =  294 +(6)M       where M is the number of minutes sfter 9:49

Subtract 294 and (1/2)M from both sides

.5  =  5.5M      divide both sides by 5.5

.5 / 5.5  = M   =  5/55 =   1/11

So....they will be together 1/11 of a minute ater 9:49 ≈ 5.45 seconds after 9:49