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can you write it i order for me 

what- oh ok thank youu

12 g / 3min   = 4 gal/min  = rate  (or slope)

y = mx + b        b = 0   (the tub starts empty)

y = 4 x              when x = 1    4 gallons

                                  x=2     8 gallons

                                  x=3    y = 12 gallons

(3x-9)(x-2) = 0

3x^2 -15x + 18 = 0      m = 9

(3x-6)(x-3) = 0

3x^2 -15x + 18 = 0      m = 9 

There is a similar question on this website: https://web2.0calc.com/questions/help-asap-plz_2

Another similar way as EP

Width = x, Height = y

2x+2y=32

xy=56

So x+y=16

Square it

x^2+y^2+2xy=256

Plug in xy=56

x^2+y^2=144

Diagonal is sqrt of that, so 12.

You are very welcome!

:P

3x +  4y   - 7    =   0        rearrange as

4y  =   -3x + 7

y  =  [ 7  - 3x] / 4   (1)

The equation of the  circle is

x^2  + y^2  =   25         sub (1)  into this for y

x^2  + [ ( 7  -3x )  / 4 ] ^2  = 25

x^2 +  [ 9x^2  - 42x + 49 ] / 16 =  25        multiply through by 16

16x^2  + 9x^2  - 42x + 49  = 400

25x^2  - 42x - 351   =  0    factor  as

(25x  - 117) ( x + 3)   =  0

Setting the second factor to  0   and solving for    produces  the  x coordinate intersection  of  pont A

So....the x coordinate of  A  =  -3   and the y coordinate of A  =  4

So  A = (-3,4)   P  = (3, 4)  and O =  (0,0)

AP =  sqrt  [ (3  - -3)^2 + ( 4  - 4)^2 ] =  sqrt  [ 6^2 + 0^2]  =  sqrt [ 36 ] =  6

So triangle AOP is isoceles with AP  =6   and  OA, OP   = 5

Using Heron's formula, the semi-perimeter   of this triangle  =  [ 5 + 5 + 6] /2=  16 / 2  =  8

Andthe area  of this triangle  =

sqrt  [8 * (8 - 6) * (8 - 5) * (8 - 5) ] =    sqrt [ 8 * 2 * 3 * 3 ]  =  sqrt [ 144]  =  12  units^2

And  using symmetry, [ AOBP ]   is twice this  =   24 units^2

First one.

If DG is at most one, we can draw a circle with radius 1 and center D.

We know the circle and triangle will share a 60 degree sector (because equilateral triangles have angle 60 degrees each).

So the area of the sector is pi/6.

The area of the entire equilateral triangle is sqrt3/4*a^2, and since a=3, 9sqrt3/4.

3.8971 is the approx. area of triangle

0.5236 is the approx. area of sector.

So 0.5236/3.8971 is around 0.134356

Or 13.4356%.

You are very welcome!

:P

You need 100   ODD numbers    1 - 199

1 3 5 7 9     and then 2-digit ones 11-99  then 3 digit ones 101-199

5                                         90                                    150 = 145 digits    (I think !)  

"Right triangle XYZ has legs of length XY=12 and YZ=6. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 12?"

The probability is 2/5.

If it did not have integer coefficients the answer would be 0.

BUT

with integer coefficients I do not know ???

If I had time and inclination I would go watch some videos on Rational Root Theorum and maybe I could figure it out.

BUT

I'd really like someone to show my how to do it here.     

Any takers?

2 l + 2w = 32

l * w = 56      l = 56/w

Sub second into first equation

112/w + 2w = 32      multiply by w

2w^2 - 32w + 112 = 0      Quadratic formula shows  w = 8+ 2 sqrt 2        then  l = 8-2sqrt2

d^2 = (8+ 2 sqrt2)^2  + (8-2 sqrt2)^2

d^2 = 144      d = 12

Let  the original   number of red marbles  =  R  and the number of  original number of  blue marbles = B

And   we know that

  R               1

_____  =    ___        cross-multiply   and we have that

R + B           5

5R  = R + B          subtract  R from both sides

4R  = B            (1)

And after 5 red marbles are added to the bag we have that

R +  5                    1

_________  =      _____      cross-multiply

R + B + 5                3

3(R +5)  =  R + B + 5

3R + 15  =  R + B + 5        subtract  R  from both sides

2R + 15  = B + 5          sub (1)  in for  B

2R + 15  =  4R  + 5        subtract  5,  2R  from both sides

10  = 2R                     divide both sides by  2

5 = R  =   the  original number of red marbles

thank you! it was correct!

\(f^{-1}(f^{-1}(6)) = 3\)

It says that it is still wrong.

Here is the Venn diagram

I would suggest a Venn diagram...I will post one in just a second...BUT you Q does not state how many dogs are in the school..... 

   Draw the three overlapping circles....

      then start where they all overlap   14 have all 3 ribbons

         then fill in the 2 circle overlaps with the info given.....

             then fill in the big circle portions with the info given.....

                 add up all of the numbers to find the number of dogs who learned a trick and got at least one ribbon

                           total 98 dogs

                                if there is more than 98 dogs in the class, the remainder learned no tricks....

To NewMember

I'd like if if you 'had it' and I a appreciate you use of 'thanks' here ........  but so far I see evidence to the contrary.

You have voted answer 1 down AGAIN

You have voted yourself up saying 'got it'.

You have NOT thanked Heureka here for linking you to his answer.

AND

You have NOT thanked Heureka for his answer on the original thread.

Seems you know how to abuse people but you don't know much about common courtesy.

I figured out the answer, but i'll post it in case anybody wants to read it.

We see that arc XZ is 180-80-35=65, so angle XYZ=(180+65)/2=122.5

l * w = 120     w = 120/l

l^2 + w^2 = 17^2 = 289

Substitute in the value of w

l^2 + 120^2/l^2 = 289     Multiply both sides by l^2

l^4 + -289 l^2 + 14400  = 0        let x = l^2

x^2 - 289x + 14400 =0       Quadratic formula reveals  x = 225  or 64

then l^2 = 225   or 64

        l = 15   or 8

         then w = 8 or 15       l x w  = 8x15=120    or   l x w = 15 x 8 = 120   

perimeter = 2l+ 2w = 2(15) + 2(8) = 46

Angle XYZ is (180 - 80 - 35) + 80/2 = 105 degrees.

Chord               >   PB = 8 

Triangle side    >   AB = 7 

Angle ABD  = 30°

Let the circle's diameter be   BD          ( Point D is 180° from point B)

Diameter         >  BD = ?                BD = AB / cos(30°)           BD = 8.083  

Chord              >  PD = ?             PD = sqrt [(BD)² - (PB)²]      PD = 1.155  

Angle      PBD > b                 cos(b) = PB / BD                       ∠ b = 8.213°   

Angle      ABP  > a                a = ∠ABD - ∠b                           ∠ a = 21.787° 

Angle      PBC  > c                c = ∠ABC - ∠a                           ∠ c = 38.213°   

PA = ?                   (PA)² = (AB)²+(PB)² - 2(AB)(PB) cos(a)         PA = 3  

PC = ?                   (PC)² = (BC)²+(PB)² - 2(BC)(PB) cos(c)          PC = 5      

Log both sx

(x+1) log 3 = log 4

(x+1) = log 4 / log 3 = 1.2619

x = .2619

then    3^(2(.2619)-1) = .5926

or another way:

3 * 3^x = 4

3^x = 4/3

3^(2x-1)  = 3^x 3^x   /3  = (4/3  * 4/3   ) / 3 =  16/9  / 3 = 16/27 = .5926

v0=5.56 m/s

vf=0  (when it hits the ground)     xo = original height   xf = final height = 0

a= -9.81 m/s^2

xf = x0 + v0 t + 1/2 a t^2

o = 108 + 5.56 t + 1/2 (-9.81) t ^2

0= 108  + 5.56 t - 4.905 t^2                          solve for t = 5.29 s