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Please note ianw11

Answer 1 is not correct, not all of it anyway.

I am pleased that you gave it a go though guest, and some of it is correct.   

The closest radical is √31 I don't know how to find the exact radical.

OQ = 4√3         ∠QOP = 15º

QR = 2*(OQ * sin∠QOP)   ==> QR = 2*(4√3 * sin15º)       

2.

Recall that a Fibonacci sequence is one in which all entries from the third onward are the sum of the two entries before it.
If the 5th and 8th entries of a Fibonacci Sequence are 19 and 79,
respectively, then what is the first entry of this sequence?

$\begin{array}{|lrcll|} \hline & a_5 &=& 19 \\ & a_8 &=& 79 \\ \hline &\mathbf{ a_6+a_7} &=& \mathbf{a_8} \\ (1) & a_6+a_7 &=& 79 \\ \hline &\mathbf{ a_5+a_6} &=& \mathbf{a_7} \\ (2)& 19+a_6 &=& a_7 \\ \hline (1)+(2): & 19+2a_6+a_7 &=& 79+a_7 \\ &19 + 2a_6 &=& 79 \\ & 2a_6 &=& 79-19 \\ &2a_6 &=& 60 \quad &| \quad : 2 \\ & \mathbf{a_6} &=& \mathbf{30} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_4+a_5 &=& a_6 \\ a_4 + 19 &=& 30 \\ \mathbf{a_4} &=& \mathbf{11} \\ \hline a_3+a_4 &=& a_5 \\ a_3 + 11 &=& 19 \\ \mathbf{a_3} &=& \mathbf{8} \\ \hline a_2+a_3 &=& a_4 \\ a_2 + 8 &=& 11 \\ \mathbf{a_2} &=& \mathbf{3} \\ \hline a_1+a_2 &=& a_3 \\ a_1 + 3 &=& 8 \\ \mathbf{a_1} &=& \mathbf{5} \\ \hline \end{array}$

Thanks Jugoslav, I get exactly the same answer (although I would have left it as an irrational number)

Here is my pic - drawn to scale

use the formula

A = 0.5ab*sinC   to find the area of triangle PQR

3 (x^2 - 28/3 x)     + 55

3 ( x - 14/3)²    - 196/3  +55

3 (x-14/3)² -31/3

     take it from here ....

A circle is centered at O  and has an area of 48pi. Let  Q and R  be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO  and triangle PQR  is equilateral, then find the area of triangle PQR.

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

OQ = OR = r = 4√3

∠QOR = 30º         ∠QOP = 15º           ∠QPR = 60º

QR = 2 (OQ * sin∠QOP) ==>         QR = 3.586301889 

Height of ΔPQR      h = sqrt[PQ² - (QR/2)²] = 3.105828542

[PQR] = 1/2(h * QR) = 5.569219383 u²

Let's say the unknown coefficient of the linear term is  b , so the quadratic expression is:

4x²  +  bx  -  24

And we want to know what  b  is.

Since we know that one of the roots is  4,  we know that if we plug in  4  for  x  into the expression, it will equal  0

So we can make the following equation:

4(4)²  +  b(4)  -  24  =  0

Can you solve the above equation for  b  ?

A pair of fair, standard dice are rolled. What is the probability the sum of the dice is 3?   

Those two dice could land in one of thirty-six combinations. 

Here they are.  The combinations that total three are purple. 

1,1   1,2   1,3   1,4   1,5   1,6  

2,1   2,2   2,3   2,4   2,5   2,6  

3,1   3,2   3,3   3,4   3,5   3,6  

4,1   4,2   4,3   4,4   4,5   4,6  

5,1   5,2   5,3   5,4   5,5   5,6  

6,1   6,2   6,3   6,4   6,5   6,6 

As you can see, there are two possibilities out of thirty-six combinations. 

So the probability is 2 out of 26  aka  2/36  aka  1/18  

_.

I'm not totally sure, but here's my guess:

Sequence A is:   30, 48, 66, 84, 102, 120, . . .

Sequence B is:   30, 12, -6, -24, -42, -60, . . .

So we can say

A₁   =   30   =   30 + 18*0

A₂   =   48   =   30 + 18*1

A₃   =   66   =   30 + 18*2

A₄   =   84   =   30 + 18*3

A_n   =   30 + 18*(n - 1)

B₁   =   30   =   30 - 18*0

B₂   =   12   =   30 - 18*1

B₃   =   -6   =   30 - 18*2

B_n   =   30 - 18*(n - 1)

And so..

The 51st term of sequence A   =   A₅₁   =   30 + 18*(51 - 1)   =   30 + 18*50   =   930

The 51st term of sequence B   =   B₅₁   =   30 - 18*(51 - 1)   =   30 - 18*50   =   -870

The absolute value of the difference between  A₅₁  and  B₅₁   =   | 930 - -870 |   =   1800

a dice has 6 sides.

6 divided by 3=2

The LCD of 6 and 2 is 6

so the answer is 2/6

Just a quick question, how did you form the equations QR = 2 (OQ * sin∠QOP) and h = sqrt[PQ2 - (QR/2)2] = 12*sqrt(3)? 

A circle is centered at O  and has an area of 48pi. Let  Q and R  be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO  and triangle PQR  is equilateral, then find the area of triangle PQR.

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

OQ = OR = r = 4√3

∠QOR = 30º         ∠QOP = 15º           ∠QPR = 60º

QR = 2 (OQ * sin∠QOP) ==>         QR = 6*sqrt(3)

Height of ΔPQR      h = sqrt[PQ2 - (QR/2)2] = 12*sqrt(3)

[PQR] = h * QR / 2 = 8*sqrt(3)

There are 5 dates that work.

[1 + 2]=3   and   [2 + 1]=3

Probability: 2 / 6^2 =1 / 18

[5 C 2 * 5^3] + [5 C 3 * 5^2] + [5 C 4 * 5^1] + [5 C 5 * 5^0]=1526 / 6^5==763 / 3888 ==19.62%. Or:

[5 C 2 * (1/6)^2 * (5/6)^3] + [5 C 3 * (1/6)^3 * (5/6)^2] + [5 C 4 * (1/6)^4 * (5/6)^1] + [5 C 5 + (1/6)^5 * (5/6)^0] =1526 / 6^5 =19.62%

Sorry, I was addressing jugoslav... his answer may be correct, but his answer is expressed as a decimal. Also jugoslav, how did you form the equation QR = 2 (OQ * sin∠QOP? 

Thanks! so it would be 5*sqrt2

Make sure you include the at least part.

It is expressed as a square root.

Can you express your answer as a sqrt please? 

6 are rolled. Probability of getting 6 is 1/6. We need at least 2. 1/6*1/6 but there are four left. We don't necessarily need 6 for these leftover 4 dice so that would by 5/6*5/6*5/6*5/6. Combining this with 1/6*1/6, that would be 1/6*1/6*5/6*5/6*5/6*5.6 which is equivalent to 0.01339591906.

Hmm... your answer is wrong Pangolin14... not really sure what went wrong though...

Using distributive property, the equation is -24-x-9+4x-12-2x. Adding/subtracting would result in b, 5x-45.