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11x^2 -44x  - 99 = 0       divide through by 11

x^2 - 4x - 9  = 0      add 9 to both sides

x^2 - 4x  =  9       take 1/2 of 4 = 2....square it = 4....add to both sides

x^2 - 4x + 4  =   9 + 4          factor the left.....simplify the right

(x - 2)^2  =  13

( x +  -2)^2  = 13

r =  -2        s  = 13

So... r + s  =   11

 2x+y ≤ 4 , x+y ≥1 , x ≥ 0 , y ≥ 0

See the following graph : https://www.desmos.com/calculator/otkx3figgt

The longest side  connects  the vertices (0,4)  and (2,0)

The length of the longest side  is   sqrt [ (2 - 0)^2 + (4 - 0)^2]  =  sqrt [ 4 + 16]  =  sqrt (20) = 2sqrt(5)

So  a  =  2  and b  = 5   

a + b  =  7

Triangle ABC has vertices A(0, 8), B(2, 0), C(8, 0)

 A horizontal line with equation y= t intersects line segment {AB}  at T and line segment {AC}  at U, forming triangle ATU with area 13.5. Compute t.

The area of the given triangle   is  (1/2)BC * height  =  (1/2)(6)(8)  = 24

Since TU  is parallel to BC.....then triangle  ATU  is similar to triangle ABC

The square of the  scale factor  * area of ABC   =  area of ATU

Scale factor ^2  =   area of ATU  / Area of ABC

Scale factor  =  √ [13.5/24]  =  3/4  

So.....the height of triangle ATU    = 3/4 the height of  triangle  ABC

So..... the line segment TU will  be located at   y  =  8 - (3/4)8  =  8 - 6  =  2   

See the pic below : 

Proof of ATU area    :  (1/2)*TU* height of 6 =   (1/2) 4.5 (6)  =  4.5 * 3  =  13.5

In triangle ABC we have (a+b+c)(a+b-c)=3ab, where a, b and c are

the sides opposite A, B and  C respectively.

What is the degree measure of \(\angle\ C=\angle\ \gamma\)  ?

\(\color{BrickRed}(a+b+c)(a+b-c)=3ab\\ a²+ab-ac+ab+b²-bc+ac+bc-c²-3ab=0\\ a²-ab+b²=c² \)

\(c²=a²+b²-\color{blue}ab\\ c²=a²+b²-\color{blue}2ab \cdot cos (\gamma)\ \color{red}(was\ wrong\ before:\ -2bc\cdot cos(\gamma))\\ 2ab\cdot cos(\gamma)=ab\\ cos(\gamma)= \frac{ab}{2ab}\\ cos(\gamma)=\frac{1}{2}\)      

was corrected by me

\(\gamma=arc\ cos\ ( \frac{1}{2})\)

\(C=\gamma=60°\)

be inapplicable:

[If it is possible to determine the ratio ( \(\large \frac{a}{c}\) \( |\ c\ge 2a\) ) from equation\( (a+b+c) (a+b-c) = 3ab\), \(\angle\ \gamma\) can also be determined with the equation \(\gamma=arc\ cos\ ( \frac{a}{2c})\).]

Danke CPhill!

  !

What specifically?

\(\frac{3-z}{z+1} \geq 1, 3-z \geq z+1\)

Now we can add z and subtract 1 to both sides, then solve and get the interval notation. Can you solve that?

You are very welcome!

:P

Edit, I think I found the domain lol.. Sorry!

Right triangle, the hypotenuse is=25 , XY=24, therefore XZ=7 (How?, 25^2-24^2=625-576=49,sqrt(49)=7) 

tan y(Angle y I assume)=opposite/adjacent = XZ/YX=7/24

so 

tan Y=7/24

....

\(\text{9 matches have 3 possible outcomes each}\\ N= 3^9 = 19683\)

....

sorry this is the link lol...

https://prnt.sc/pj5m80

x^2 + x - 5  = 0

Let a  = 1      b  = 1    c   = -5

Using the quadratic formula  we have that

x =     -1 ±√ [ 1^2 - 4* 1 * -5 ]       =        -1  ± √ [1 + 20 ]            -1  ± √21

         ___________________              _____________    =     ________

                     2 * 1                                           2                              2

So  D, F    are the correct answers

Let the unknown side = S

By the triangle inequality

S < 3 + 5

S < 8

S + 3 > 5

S > 2

And

S + 5 < 3

S > -2

But  taking the most  restrictive interval

2 < S < 8

But since the triangle is scalene.....then the possible lengths for S  are  4, 6, and 7 (cm)

The problem with reducing the count by dropping one of the letters is it reduces the sample space. Each letter switches place with every other letter multiple times (half of all counts) as the entropy decreases to zero. Each E will switch places with every other letter and all of these events are unique, except for when the E’s switch with each other. 

Dropping the count by one (1) and removing the E, means the one of the E’s will not switch with any other letter. By keeping the E and its counts with all letters, it remains only to factor out the redundancies caused by switches with its twin.

This brings up an interesting point: There are times when it should not be factored out.

I note the title to your post is “Scrabble probability questions” (Great descriptive title. It’s one of the reasons I answered your post.) There is no probability question here, but counting is a precursor to probability. When counting for probability the redundancies are always counted.

Consider these questions: John randomized his seven (7) tiles and creates a word.

What’s the probability the word starts with Z?   Answer: (1/7)

What’s the probability the word starts with E? Answer: (2/7)

It’s easy to see that it’s counted here, and it’s always counted in combination and permutation problems related to probability. 

GA

thank you

thank you

thank you

thank you

thank you

You should use the formula. This will help to give you an “intuitive” understanding of how the TVM works. Save the online financial calculators for checking your work. If you just use financial calculators or software to solve these, it will alter your DNA, and you’ll become a Blarney Banker.

\(\text {Here is the formula ...}\\ \text { }\\ \dfrac{FV*\left(\dfrac{r}{n}\right)} {\left(1+\dfrac{r}{n}\right)^{nt} -1} =D\\ \text{ }\\ \small \dfrac{3000000*\left(\dfrac{0.085}{12}\right)} {\left(1+\dfrac{0.085}{12}\right)^{(12*40)} -1} = 742.82\\ \text{ }\\ \small \text {A monthly deposit of 742.82 dollars for 40 years and the accrued compounded interest will have a future value of 3,000,000 dollars. }\\\)

Paste this into the forum’s calculator:

f=3000000;r=0.085;n=12;t=40; f*(r/n)/ ((1+r/n)^(n*t) -1)

GA

We have 20 people of team A ,20 people of team B and 20 double rooms .How we can put the peaople so that in the rooms we only have people from the same team.

How many ways can 10 rooms be chosen from 20 rooms

20C10 = 184756

Team A can have these or they can have the other set. So that is 2 ways.

Now how many ways can I pair off 20 people?

Person 1 can go with anyone, 19 to chose from

Now there are 18 people left

Person 3 can go with anyone 17 to chose from

so there will be 16 left

and so on.

So that seem to be   19*17*15*13*11*9*7*5*3*1 = = 654729075

I really do not know if that is correct.

So I have

184756*2*654729075*654729075 158398768771746207705000 = 1.5839876877174621 * 10^23

Mmm that is a lot!

That is my best guess. (For the moment)

Thanks for the reply, I appreciate it a lot! It's very clear and straightforward. I don't understand one thing though. You say that the two E's are indistinguishable, so shouldn't it be (6!/2!) instead then? Sorry if this is a stupid question, I'm kind of confused.

You must deposit $1,149.26 at the end of every 3 months for 5 years or 5 x 4 = 20 quarters in order to have $25,000 in your investment account.

 [$1,149.26 x 20] = $ 22,985.20 - This is total principal in deposits.

$25,000 - $22,985.20 = $2,014.80 - This is total interest earned on the investment.

The digits 1 up to 4 are written on a spinning top. Ella spins the top five times. This way she'll get a number of five digits, like 42313 or 13421

a. How many different numbers can Ella get? 45 = 1024 numbers

4^5=1024 different possibilities     Correct

b. How many different numbers can she get if the number starts with an even digit and then alternates with an odd digit and then an even digit. 25 = 32

even, odd, even, odd, even,    2^5=32       correct

c. How many different numbers can she get if she spins two times a 2, two times a 4 and then one other (1 or 3) digit?

2,2,4,4,  (1 other 2 to chose from)

22441,   or  22443      in any order

5!/(2*2)   each

30 each

60 possibilities    (I'm pretty sure)