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are you asking how many bingo cards have a top left to bottom right diagonal whose values are strictly increasing?

The graph "kisses" the x axis at  x = 1   and goes through the x axis at x = -2

Therefore....we have that

y  =  (x + 2) (x - 1)^2

When x = 0, then "c"  =  2

So....we have that

( x + 2)(x-1)^2  = 2

(x + 2) ( x^2 - 2x + 1)  = 2

x^3 +2x^2 -2x^2 -4x + x+ 2  = 2

x^3 - 3x = 0

x( x^2 - 3) = 0       set both factors to 0 and solve for x

x = 0        and     x^2 - 3  = 0  ⇒  x^2 = 3 ⇒     x  = ± √3

Here's a graph :  https://www.desmos.com/calculator/co5bdggyfq

Thank you!

Btw the ⊂ symbol means just a subset 

If the 5th and the 8th entries of a Fibonacci Sequence are 19 and 79, respectively, then what is the first entry of this sequence?

List of Fibonacci numbers with \(a_0 = 0\) and \(a_1 = 1\):

\(\small{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \ldots& F_{-2} & F_{-1} & F_0 & F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 &F_9 &F_{10} &F_{11} &F_{12} &F_{13} &F_{14} &F_{15}&\ldots \\ \hline \ldots& -1 & 1 & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610&\ldots \\ \hline \end{array} }\)

In general:

\(\begin{array}{|rcll|} \hline a_n &=& F_na_1+F_{n-1}a_0 \quad | \quad \text{with } a_0=0 \text{ and } a_1 = 1,\quad \text{yield}\quad a_n =F_n \\ \hline \end{array} \)

\(\mathbf{a_1=\ ?}\)

\(\begin{array}{|lrcll|} \hline n=5: & F_5a_1+F_4a_0 &=& a_5 \quad | \quad a_5 = 19 \\ n=8: & F_8a_1+F_7a_0 &=& a_8 \quad | \quad a_8 = 79 \\ \hline & \mathbf{F_5a_1+F_4a_0} &=& \mathbf{ 19 } \qquad (1) \\ & a_0 &=& \dfrac{19-F_5a_1}{F_4} \quad | \quad F_4=3,\ F_5=5 \\ & a_0 &=& \dfrac{19-5a_1}{3} \\\\ & \mathbf{F_8a_1+F_7a_0} &=& \mathbf{79} \qquad (2) \\ & F_8a_1+F_7\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad F_7=13,\ F_8=21 \\ & 21a_1+13\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad \cdot 3 \\ & 63a_1+13\cdot ( 19-5a_1 ) &=& 237 \\ & 63a_1+247 - 65a_1 &=& 237 \\ & -2a_1 &=& -10 \\ & 2a_1 &=& 10 \\ & \mathbf{ a_1 } &=& \mathbf{ 5 } \\ \hline \end{array}\)

The first entry of this sequence is 5

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).
Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array} \)

\(\mathbf{\text{vieta:}}\)

For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0 \)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:

   \(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array} \)

The answer is 8.

Given    \(f(x)=3x^2+10x-8\)    and    \(g(x)=3x^2-2x\)

What is \((\frac{f}{g})(x)\)   ?

--------------------------

\((\frac{f}{g})(x)\,=\,\frac{f(x)}{g(x)}\)

                                              Substitute  \(3x^2+10x-8\)  in for  \(f(x)\)  and substitute  \(3x^2-2x\)  in for  \(g(x)\)

\((\frac{f}{g})(x)\,=\,\frac{3x^2+10x-8}{3x^2-2x}\)

                                              Split  10x  into two terms such that their coefficients add to  10  and multiply to  -24

\((\frac{f}{g})(x)\,=\,\frac{3x^2+12x-2x-8}{3x^2-2x}\)

                                              Factor the numerator and denominator.

\((\frac{f}{g})(x)\,=\,\frac{3x(x+4)-2(x+4)}{3x^2-2x}\\~\\ (\frac{f}{g})(x)\,=\,\frac{(x+4)(3x-2)}{x(3x-2)} \)

                                              Cancel the common factor of  (3x -2)  and note that  3x - 2 ≠ 0 , that is,  x  ≠  2/3

\( (\frac{f}{g})(x)\,=\,\frac{x+4}{x}\qquad\text{and}\qquad x\neq\frac23\)

Since  x  is in the denominator we can also note that  x ≠ 0

(a) I want the midpoint value of the interval for the point estimate, so \(\frac{0.31+0.52}{2} = 0.415\)

(b) The poll's margin of error can be described as how far either end of the interval is from the point estimate, so \(|0.52-0.415| = 0.105\)

(c) estimate \(= \hat{p} * n = 0.415(140371) = 58253.965\)

A. \(\hat{p} = \frac{56}{80} = 0.7\)

B. estimated \(= \hat{p}*n = 0.7(138246) = 96772.2\)

Well, since this is inverse sine of a sine it cancels out to return to the given angle of \(\frac{3\pi}{8}\).

But, we should always check where stuff is defined with inverse trig functions and make sure \(\frac{3\pi}{8}\) is within it. The range of \(sin^{-1}\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\), and \(\frac{3\pi}{8}<\frac{\pi}{2}\), so the answer is \(\frac{3\pi}{8}\). 

A general Fibonacci sequence is defined as: \(a_n = a_{n-1} + a_{n-2}\)

We can use what was given, that \(a_5 = 15 \) and \(a_8 = 79\) to solve for \(a_1\)

\(a_6 + a_7 = a_8 \rightarrow a_6 + a_7 = 79\)

\(a_5 + a_6 = a_7 \rightarrow 19 + a_6 = a_7\)

Substituting for \(a_7 \) in the top equation , we can see \(a_6 + (a_6 + 19) = 79\)

\(2a_6 = 60 \rightarrow a_6 = 30\)

This part gets a lil repetitive

\(a_{4} + {19} = {30} \rightarrow a_{4} = {11}\)

\(a_{3} + {11} = {19} \rightarrow a_{3} = {8}\)

\(a_{2} + {8} = {11} \rightarrow a_{2} = {3}\)

\(a_{1} + {3} = {8} \rightarrow a_{1} = {5}\)

Therefore, the first entry of this sequence is 5.

If  A = {2, {4, 5}, 4}, true or false:

Our answers are different on 1 and 3.

Also...I don't know if the ⊂ symbol means a proper subset or just a subset, but it doesn't really matter for these questions. 

Thank you!

Just to reformat your question for my aesthetic quick;

Simplify \((x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)\) to the form \(0 = Ax^{2}+Bx+C\) where A, B, and C are positive integers with a greatest common divisor of 1. What is \(A+B+C\)?

\((x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)\)

\((2x^2+3x+2x+3)(3x+4) = 6x^3-18x^2+5x-15\)

\((2x^2+5x+3)(3x+4)=6x^3-18x^2+5x-15\)

\(6x^3+8x^2+15x^2+20x+9x+12=6x^3-18x^2+5x-15\)

\(6x^3+23x^2+29x+12=6x^3-18x^2+5x-15\)

\(41x^2+24x+27=0\)

\(Ax^2 + Bx + C = 0\)

\(A = 41, B = 24, C = 27\)

\(A + B + C = 41+24+27=92\)

Thank you, sorry for not searching farther ;)

2.

What is the parent function and what general shape does the parent function have?

The parent function is  f(x)  =  |x|

Its general shape is a "V" shape.

Explain all of the parameters and what they do to change the graph.

The graph of  f(x)  =  -3|x + 1| - 2  is the graph of  f(x)  =  |x|

shifted to the left by  1  unit,

stretched vertically by a factor of  3,

flipped over the x-axis,

and shifted down by  2  units.

1  is added to x, which shifts the graph to the left by  1  unit.

3  is multiplied by  |x + 1|  which stretches the graph vertically by a factor of  3 .

-1  is multiplied by  3|x + 1|  which flips the graph over the  x-axis.

-2  is added to -3|x + 1|  which shifts the graph down by  2  units.

P(B|A)=P(B)

Since they are independent......the P(A)  has no effect on the probabilty of B

1.

Graph the function  \(f(x)=\begin{cases} x+6 & x\leq -4 \\ (x+2)^2-3 & -4< x\leq 1\\ -|x-4|+5 & x>1 \end{cases} \)

Here's the graph of the function on desmos:

We can just label the points  as  (4, 0) (3,5) (-2, 4) and (-1, -1)

The distance between (3, 5)  and   (-1, -1)  will be one diagonal   

So....the distance is

sqrt [ (-1-3)^2 + ( (-1-5)^2 ] = sqrt ( (-4)^2 + (-6)^2 ]  = sqrt [ 16 + 36 ]=  sqrt [52 ]  =

sqrt(4) * sqrt(13)  =     2sqrt (13)  units

2x + 4y  = 6     to fid the y intercept, let x  = 0

2(0) + 4y  = 6

4y  = 6    divide both sides by 4

y = 6/4  =  3/2  ⇒  y intercept

(x+y)^2=x^2+y^2+2xy.

x^2+y^2+2(13)=78, x^2+y^2=52.

(x-y)^2=x^2+y^2-2ab, so 52-2(13)=52-26=26-answer.

Yes, thanks.

a.  On the curve....put 71  in the middle   and the three deviations (the vertical segments to the base of the curve ) to the right will be  77, 83  and 89

And the 3 deviations to the left - starting from the mean and  working to the left-  will be  65, 59  and 53

b.   z score  =  [ Angelic'a grade - Mean ] / Standard Deviation =  [ 76 - 71 ] / 6  =  5/6 = . 83

c. Looking at a z score table.....   .83  equates to ≈  .7967   =  79.67%

This means that she scored highr than about 80%  of the students

d.  About   20% scored higher....so.....20 (185) =   37 students