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$\frac{9 \cdot 3 + 2 \cdot 9}{11}=\frac{9(3+2)}{11}=\frac{45}{11}$

Okay, I actually haven't solved the problem but I quickly noticed some things.

First, we can ignore all horizontal lines (we can't move sideways so thinking about moving left or right isn't exactly correct).

Second, due to ignoring the horizontal lines, we can limit our problem to a rhombus shape (forgive me I don't know how to create diagrams).

Now instead of thinking it as a rhombus shape imagine it as a square (or a 4x4 grid as shown)

Take one corner as the entrance and the opposite corner as the exit and now it's become an easier path problem.

Hope this makes sense.

Here's a similar one by CPhill, change the values and plug in! https://web2.0calc.com/questions/suppose-we-have-a-bag-with-10-slips-of-paper-in-it

(a - b)   / (a + b)   =  3/5

5 (a - b)  =  3(a +b)

5a  - 5b  =  3a  + 3b

5a - 3a  =  3b + 5b

2a  =  8b

a/b  =   8/2   =  4

ab    =   192

bc   =    360

cd   =    750

Let  a =  8      b =  24    c  =  15      d  =  50              sum =   97

Let  a = 16     b= 12      c =    30     d =   25              sum =   83

Thanks Wsai, here is another way to do it

Divisibility rule for 72 : Divisible by 8 and 9

\(\overline{a679b} \) = 10000a+6000+700+90+b

so we have two equations:

(790+b)/8 is an integer

a + 6+7+9+b is divisible by 9

so a=3 and b=2

Check

(790+2)/8 = 99 :)

(3 + 6  + 7 + 9 + 2)/9 = 3:)

Here you go guest:

|x-2| + |x-7| - |x-8| = 5

x=-4, x=22/3

Triangles   ABD    and   CBD  will  be  under  the  same  height

So     [ABD]  =    (8 / ( 3 + 8) )  *  120   =   (8)(120)  /11  = 960/11    ≈  87.27

part a is wrong but the rest are corrcect

I see, thanks!

There are this many sequences of rolls to obtain a product of 300:

12556 , 12565 , 12655 , 13455 , 13545 , 13554 , 14355 , 14535 , 14553 , 15256 , 15265 , 15345 , 15354 , 15435 , 15453 , 15526 , 15534 , 15543 , 15562 , 15625 , 15652 , 16255 , 16525 , 16552 , 21556 , 21565 , 21655 , 22355 , 22535 , 22553 , 23255 , 23525 , 23552 , 25156 , 25165 , 25235 , 25253 , 25325 , 25352 , 25516 , 25523 , 25532 , 25561 , 25615 , 25651 , 26155 , 26515 , 26551 , 31455 , 31545 , 31554 , 32255 , 32525 , 32552 , 34155 , 34515 , 34551 , 35145 , 35154 , 35225 , 35252 , 35415 , 35451 , 35514 , 35522 , 35541 , 41355 , 41535 , 41553 , 43155 , 43515 , 43551 , 45135 , 45153 , 45315 , 45351 , 45513 , 45531 , 51256 , 51265 , 51345 , 51354 , 51435 , 51453 , 51526 , 51534 , 51543 , 51562 , 51625 , 51652 , 52156 , 52165 , 52235 , 52253 , 52325 , 52352 , 52516 , 52523 , 52532 , 52561 , 52615 , 52651 , 53145 , 53154 , 53225 , 53252 , 53415 , 53451 , 53514 , 53522 , 53541 , 54135 , 54153 , 54315 , 54351 , 54513 , 54531 , 55126 , 55134 , 55143 , 55162 , 55216 , 55223 , 55232 , 55261 , 55314 , 55322 , 55341 , 55413 , 55431 , 55612 , 55621 , 56125 , 56152 , 56215 , 56251 , 56512 , 56521 , 61255 , 61525 , 61552 , 62155 , 62515 , 62551 , 65125 , 65152 , 65215 , 65251 , 65512 , 65521 , Total =  150 such sequences.

If we want a quadratic equation to have 1 root, the discriminant must be 0.

so, b² - 4ac = 0

a=3

b=4

c = ?

4² - 4(3)(c) = 0

16-12c = 0

-12c=-16

c=16/12

c=4/3

In order for the domain to not work, the denominator.

x²-5x-4=0

\(x = {5 \pm \sqrt{25+16} \over 2}\)

A = (5+√41)/2

B = (5-√41/2)

A+B = (5+√41+5-√41)/2

A+B = 5

a)   3(8/10)  +  9(2/10)  =  24/10  + 18/10   =   42/10  =   4.2

b)  3(8/11)  + 9 (3/11)  =  24/11   +  27 /11   =  51/11  ≈   4.63

c)  3 ( 8/12)  + 9(4/12)  =  24/12  + 36/12  =   60/12  =   5

We have two tasks.....choose  the correct  white balls  and choose the correct SuperBall

Number  of  possible  sets   of  C(10,3)   = 120.....only one of  these  will win

So    1/120

And  the probability of  choosing  the  correct SuperBall  =  1/10

So

(1/120)  *  (1/10)  =  1  /1200

Note: make sure to know sin/cos angle addition/subtraction identities, sin/cos double angle identities, the Pythagorean identity, and the proofs for each one of them.

Using the sine addition rule,

\(\sin(3\theta)=\sin(\theta+2\theta)=\sin(\theta)\cos(2\theta)+\sin(2\theta)\cos(\theta)\)

And then apply the double angle identities:

\(\sin(\theta)(1-2\sin^2\theta)+2\sin(\theta )\cos(\theta)\cos(\theta)\\ =\sin(\theta)(1-2\sin^2\theta+2\cos^2(\theta))\)

Note that by the Pythagorean identity,

\(\sin^2(\theta)+\cos^2(\theta)=1\\2\sin^2(\theta)+2\cos^2(\theta)=2\\-2\sin^2(\theta)+2\cos^2(\theta)=2-4\sin^2(\theta)\)

Substituting the value above, we get:

\(\sin(\theta)(1+2-4\sin^2(\theta))\\=3\sin(\theta)-4\sin^3(\theta) \blacksquare\)

A very similar proof can be used to prove the latter.

Thank you. I understand now. 

D is on side AC of Triangle ABC. We know [ABC] = 120, AD = 8, DC = 3. What is [ABD]?

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

Let ΔABC be the right triangle.

If AD = 8, DC = 3, and [ABC] = 120, then AB = 21 ⁹/₁₁

[ABD] = 1/2(AD * AB) = 87 ³/₁₁ square units

x^2  -  15x  +  t  =   0

Note  that  we  have  these possibilities  for  t

(x - 14) ( x - 1)     t =   14

(x - 13) ( x - 2)     t  =  26

(x - 12) ( x - 3)     t =   36

(x - 11) ( x - 4)     t =   44

(x - 10) (x - 5)     t =    50

(x - 9) ( x - 6)      t =    54

(x - 8) ( x - 7)      t =    56

Avg  =   ( 14 + 26 + 36 + 44 + 50 + 54 + 56)  / 7  =  280  / 7  =    40

3a  +  4b   =  7       multiply  through  by 2

6a   +  8b   =  14

And note  that  the second equation simplifies to

6a  + 8b  =  k

Which....if we have infinite solutions implies  that  k =   14

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

(x + iy)^3   =   -74  + ki

x^3 + 3x^2*iy  +  3x * (iy)^2  +  ( iy)^3  = -74  + ki

(1)^3  + 3*yi  +  3y^2i^2   + (yi)^3  =  -74  +  ki

1  +  3yi   - 3y^2  - y^3i  =  -74  + ki

  - 3y^2  +  (3yi  - y^3i)  =  -75  + ( ki  )

Equate 

-3y^2  =  -75

y^2 =  -75/-3

y^2 =  25

y  =  5   or y =  -5

If  y   = 5....then                                  If  y =  -5......then

3(5)i - (5^3)i    =  ki                                3 (-5)i  - (-5)^3 i  =  ki

15i - 125i = ki                                          -15i  + 125i  =  ki

-110 i =  ki                                                110i  =  ki

-110  =  k                                                  110    =  k

The value of b is 18.

(2a - - 2)   /( 6 - - a)  =   7

(2a  + 2)  / ( 6 + a)   = 7

2a + 2   =   7 (6 + a)

2a  + 2  =  42   + 7a

2a  - 7a    = 42  - 2

-5a  =  40

a =   40  / -5     =  -8