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\(5\) is correct! Great job CPhill I love the way you solved this. 

-811   mod 24

One way of doing this  :

Thinking "backwards".....note  that  811/24  =   33.79......

Round this up to the  next integer =  34 

And note that   34 *24  = 816

The  residue  =   816 - 811   =   5

See the following  :

Note that this boils down to  finding  the area  of  the quarter circle  with a center at the origin and with a radius of 1 within a square with a side of 2

Area of the quarter circle  = pi (1)^2 / 4   =   pi/4

Area of the square  = 2^2   = 4

Probability =    area of quarter circle  / area of square  =   (pi/4)  / 4  =   pi / 16 

We can use something known as Heron's Formula to solve this

First.....find the semi-perimeter, S   =   [ 13  + 17  + 12sqrt (3)  ] / 2  =  15 + 6sqrt (3)

Area   =   sqrt  [  S  * ( S -A)  *  (S - B)  * ( S - C) ]  .....  [  A,B, C  are the triangle sides ]

Area  = sqrt  [  (15 + 6sqrt (3) )   *   ( 15 + 6sqrt (3)  - 13) *  ( 15 + 6sqrt(3) - 17)  *  (15 + 6sqrt (3) - 12sqrt(3)) ]  =

sqrt  [ (15 + 6sqrt (3))  * (2 + 6sqrt(3)) * (6sqrt (3) - 2)  *  (15 - 6sqrt (3)) ] =

sqrt  [ ( 6sqrt(3) + 2) * ( 6sqrt (3) - 2) * ( 15 + 6sqrt (3)) * ( 15 -6sqrt (3) ) ]  =

sqrt [  (36*3 - 4)  * (225- 36*3) ]  =

sqrt  [ 108 - 4) * ( 225 - 108) ] =

sqrt  [ 104  * 117]  =

sqrt  [ 26 * 4  * 13 * 9] =

sqrt [ 13 * 2  * 4  * 13  * 9 ]  =

sqrt [ 13^2  * 2^2 * 3*2  * 2  ] =

13 * 2 * 3  * sqrt (2)  =

78 sqrt (2)  units^2

We can choose any of the 5 red shirts

Any of the 4 pairs of pants

Any of the (7 green  + 8 red) hats  = 15   (we canot choose  the same  color of red as the shirt we chose)

So...the total number of outfits  =   5 * 4 * 15  =  300

Hello Guest!

Since your week is $5$ days, and there is a $40\%$ chance of sun, we can say that "we expect" $2$ days to have sun. Now, since we have a $20\%$ chance of $4$ inches of rain, we can "expect" $1$ day to have $4$ inches of rain. Now, we have a $40\%$ chance of $10$ inches of rain, hence the last $2$ days we "expect" to have $10$ inches of rain. Now we have $4+10+10=24$ inches of rain from Monday to Friday. This is my take on this problem. Please tell me if this is wrong or does not satisy your problem so I can review and revise. 

3-2/3x=2/5  and without decimal

Hello Guest!          | use the google translator once

\(3-\frac{2}{3}x=\frac{2}{5}\)         | - 3  beide Seiten

\(-\frac{2}{3}x=\frac{2}{5}-3\)     | mal - 1

\(\frac{2}{3}x=3-\frac{2}{5}\)        | rechts gemeinsamen Bruch bilden

\(\frac{2}{3}x=\frac{3\cdot 5{\color{Red}-2}}{5}\)        | Zähler ausrechnen   ( -2 war +2, danke Alan)

\(\frac{2}{3}x=\frac{\color{Red}13}{5}\)            | mal \(\frac{3}{2}\)   beide Seiten

\(x=\frac{{\color{Red}13}\cdot 3}{5\cdot 2}\)            | multiplizieren   

\(x=\frac{\color{Red}39}{10}\)              | rechts durch 10 mit Rest

\(x=3\frac{9}{10}\)

  !

Another hint.  Factor the denominator x² + 7x – 30  to  (x + 10)(x – 3)  then read Melody's hint again. 

_.

Ok Ginger, not a problem.   

Hi Melody,

I figured out why you couldn’t follow my logic.  My solution has a shovel-full of bullshit 🐃💩 mixed in with the logic and math.  I didn’t see or smell this until I started constructing another solution;  I’m still not sure where it came from.  It’s Spooky!  (This may be an example of Nauseated’s Spooky Dumbness at a Distance.) 

My new solution gives (36.364%) as the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence. 

I will present the solution with coherent details after I triple check it for logic and counting errors.

GA

Can somebody help him because me too I need the answer

Think about

\(\mathbf{A^7}\mathbf{x}= \mathbf{A^6}(\mathbf{A}\mathbf{x})\)

consider (5x-4)^2

no matter what real value x has, this expression will be some real number squared.

When you square ANY real number the outcome CANNOT be NEGATIVE.

So thee is no way that this expression can be less than 0.

If x=4/5  then the expression will be 0

Consider if x=0 (which is less than 4/5)

(5*0-4)^2 = (-4)^2 = 16     which is definitely not less than 0.

Thanks!

You did something clever right here:> https://web2.0calc.com/questions/pre-cal-need-help

through: (-4,2), perp. to y=4/7x+5

Hello Guest!

\(f_1(x):\\ y=\frac{4}{7}x+5\\ m_1=\frac{4}{7}\)

\(f_2(x):\\ m_2=-\frac{1}{m_1}=-\frac{1}{\frac{4}{7}}\\ \color{blue}m_2=-\frac{7}{4}\\ \color{BrickRed}P(-4,2)\)

\(y=m(x-x_P)+y_P \)   (\(point-direction\ equation\))

\(y=-\frac{7}{4}(x-(-4))+2\\ y=-\frac{7}{4}x-7+2\)

\(y=-\frac{7}{4}x-5\)

  !

urw, Phill!

Important: When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

∠SOT = xº         ∠ROB = (4x)º            ∠RTB = 28º

 ∠RTB = 1/2 (∠ROB - ∠SOT)     

28º = 1/2 (4x - x)            x = 18.6666666

Arc RS = 180 - 5x = 86.6666666º

Hes now the top on my leader board

Thx, hugoslav  !!!!

Melody  has set up a partial fraction problem......thanks, Melody!!!

It may not be entirely clear  what to do after her last step

Note  that if we multiply through by the common denominator of  (x-3) (c-5)  we get that

2x+ 7 = A(x-5)  + B (x - 3)

2x + 7 = (A + B)x  -5A  - 3B 

This implies  that we have this system  :

A + B =  2

-5A - 3B = 7

Manipulating the first  we have that  B  = 2  - A

So....we  have  that

-5A - 3 ( 2 -A)  = 7

-5A  -6 + 3A  = 7

-2A  =  13

A = -13/2

And B  = 2 -  - (13/2)  =  17/2

AF = sinB * c

BF = AF / cosB 

CF = sqrt(b² - AF²)

BC = BF + CF

When you have all 3 sides, use the law of cosines to calculate the other 2 angles.