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^{Who are you lecturing as "Mr. BB", you monkey face? How do know what the questioner meant?}

I’m lecturing Mr. BB(₄) --the Bullshit Bug, a genetic variant of the dung beetle that, instead of rolling dung into a ball, it spreads it allover the forum. This is an accurate description of what you usually do.

Determine wheter the given statement is true or false.

The statement is wrong.
F starts at the beginning of H and goes diagonally down to the right
to the end of G.

Greatings 

   !

Mr. BB,  you are confusing prime factors with factors. There are at least two other questions where you have conflated the two. They are not the same thing. I corrected your screw ups, but I didn’t troll you for it, so we’re dealing with it again.

Write this down on a sticky-note and stick it on your computer screen.  Prime factors and factors (divisors) are NOT the same thing!

GA

r =      5 / [ 1 + cos θ]

Note :  r = √ [ x^2 + y^2]       and    cos θ  =  x / r   

So we have

r = 5 / [ 1 + x/r ]

r [ 1 + x / r]  =   5

r  + x  = 5

 √ [ x^2 + y^2]  +  x  = 5      subtract  x from both sides

 √ [ x^2 + y^2]   =  5  - x      square both sides

x^2 + y^2 =  25  - 10x  + x^2       subtract x^2 from  both sides

y^2  = 25 - 10x

Hi,

Look here.

Please help someone!!!

\(2\frac{7}{9}\) would be appropriately classified as a mixed number, a number with an integral part and a fractional part.

Guest, if you want an answer you should probably include the data table in your problem description.

There is no data table in your problem description.

I think this can be solved by a graph as well, Coldplay...here :  https://www.desmos.com/calculator/myszb3kghi

The total possible area for (x, y) is a 3 by 3 square = 9 units^2

The feasible region of  (x, y)  values occur inside the boundary points (0,1), (1,0), (2,3), (3,2) and (3,3)

We have an  upper rigtt triangle with an area of (1/2) (product of the leg lengths)  = (1/2) (1) (1)  = 1/2

And we have a rectangle of length  √ [ (3 - 1)^2 + (2 - 0)^2 ] = √ [2^2 + 2^2] = √8

And a height of  √[ (2-3)^2 + (3 - 2)^2] = √ [1^2 + 1^2 ] = √2

So...the area of this rectangle is  √8 * √2 = √16  = 4

So....the toal area of the feasible region is  [ 4 + 1/2]   =  4.5 units^2

So...the probability that a triangle exists  is   4.5 / 9   =  1/2

See the solution here: https://web2.0calc.com/questions/help-please_46127

Omi is probably correct, I have not checked either answer. 

Anyway guest, you have the method.

Thanks Omi :)

https://www.biography.com/people/leonhard-euler-21342391

The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12 , then what are the $x$-intercepts of the graph of f? Explain.

Thanks for responding to my answer.  The first one I mean.  

Mmm lets see.

\(a+c=-10\\ 4a+c=2\\ so\\ 3a=12\\ a=4\\ c=-10-4=-14\\\)

So we have

\(f(x)=ax^3+bx^2+cx+d\\ f(x)=ax^3+5x^2+cx\\ f(x)=4x^3+5x^2-14x\\ f(x)=x(4x^2+5x-14)\\ F(x)=0\;\; when \;\; x(4x^2+5x-14)=0\\ x=0\;\;or\;\;4x^2+5x-14=0\\ x=0\;\;or\;\; x=\frac{-5\pm \sqrt{25+224}}{8}\\ x=0\;\;or\;\; x=\frac{-5\pm \sqrt{249}}{8}\\ \)

If there are no careless errors in my working then these are the x intercepts.

You should check my answer by substituting the answers in for x and seeing if the answer for f(x) is zero.

Hi Ginger, it is nice to see you back again. :)

Find all solutions of the equation(answers)

|x^2 - 14x + 29| = 4

\(\begin{array}{|lrclcrcl|} \hline & |x^2 - 14x + 29| &=& 4 \\ \Rightarrow & |(x-7)^2-20| &=& 4 \quad & | \quad \text{substitute} \quad u = (x-7)^2 \\\\ & |u-20| &=& 4 \\ & u_1-20 &=& 4 &\text{ or }& u_2-20 &=& -4 \\ & u_1 &=& 24 && u_2 &=& 16 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline u &=& (x-7)^2 \\ \pm\sqrt{u} &=& x-7 \\ 7 \pm\sqrt{u} &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{7 \pm\sqrt{u}} \\ \hline \end{array}\)

Solutions:

\(\begin{array}{|rclcl|} \hline x_1 &=& 7+\sqrt{u_1} = 7 + \sqrt{24} &=& 7+2\sqrt{6} \\ x_2 &=& 7-\sqrt{u_1} = 7 - \sqrt{24} &=& 7-2\sqrt{6} \\ x_3 &=& 7+\sqrt{u_2} = 7 + \sqrt{16} &=& 11 \\ x_4 &=& 7-\sqrt{u_2} = 7 - \sqrt{16} &=& 3 \\ \hline \end{array}\)

Proof:

\(\begin{array}{|rcll|} \hline |(7+2\sqrt{6})^2 - 14\cdot (7+2\sqrt{6}) + 29| &=& 4 \\ |11.8989794856^2 - 14\cdot 11.8989794856 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |(7-2\sqrt{6})^2 - 14\cdot (7-2\sqrt{6}) + 29| &=& 4 \\ |2.10102051443^2 - 14\cdot 2.10102051443 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |11^2 - 14\cdot 11 + 29| &=& 4 \\ |121 - 154 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |3^2 - 14\cdot 3 + 29| &=& 4 \\ |9 - 42 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\ \hline \end{array}\)

Solution:

\( \text {Three of the six will remain at the game until its conclusion, this leaves three (3) }\\ \text {with a 40% probability of remaining at the game until its conclusion.}\\ \text {Find the probability that two (2) or more of these three (3) will remain. }\\ \text {To find this probability, calculate the Binomial CDF for } \large \rho \small (X \geq 2, 0.40)\\ \large \binom{3}{2}*0.4^2*0.6^1 + \binom{3}{3}*0.4^3 = 0.352\\ \;\\ \text {The probability is } 35.2\%\\ \)

GA

3)
(1/4)x^2 + 12=0

\(\begin{array}{|rcll|} \hline (1/4)x^2 + 12 &=& 0 \quad | \quad \cdot 4 \\ x^2 + 48 &=& 0 \quad | \quad -48 \\ x^2 &=& -48 \\ x^2 &=& (-1)\cdot 16 \cdot 3 \\ x &=& \pm \sqrt{(-1)\cdot 16 \cdot 3} \\ x &=&\pm\sqrt{16}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 4\sqrt{3}\cdot i \\ \hline \end{array}\)

2)
6x^2 = -126

\(\begin{array}{|rcll|} \hline 6x^2 &=& -126 \quad | \quad : 6 \\ x^2 &=& -21 \\ x^2 &=& (-1)\cdot 21 \\ x &=& \pm\sqrt{(-1)\cdot 21} \\ x &=&\pm \sqrt{21}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm \sqrt{21}\cdot i \\ \hline \end{array}\)

1)

3x^2 + 81=0

\(\begin{array}{|rcll|} \hline 3x^2 + 81 &=& 0 \quad | \quad : 3 \\ x^2 + 27 &=& 0 \quad | \quad -27 \\ x^2 &=& -27 \\ x^2 &=& (-1)\cdot 9 \cdot 3 \\ x &=& \pm\sqrt{(-1)\cdot 9 \cdot 3} \\ x &=&\pm\sqrt{9}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 3\sqrt{3}\cdot i \\ \hline \end{array}\)