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Measure of angle QTR  =  (1/2)arc QR  = (1/2)(110)  = 55°

Measure of angle RQS  = (1/2) arc RS  = (1/2)(94)  = 47°

Measure of arc TS  =  [ note that arc QT  =  2 m angle ORT  =  2*27  =  54°

So  measure of arc TS  =

360 - arc QR - arc RS - arc  QT  =

360 - 110 - 94 - 54  = 

102°

Measure of angle TRS  =  (1/2)arc TS  =  (1/2)(02)  = 51°

Measure of angle QSR  =  measure of angle QTR  = 55°

Let's see if we can get a larger answer.....

Mean  = 6

Mode = 5

Median = 4

 1  2   3  4  5  5  22

Yes!!!.....you are correct,  Nirvana.....we CAN do better than 15  !!!!

Good, job!!!!

Matrix P  =                                               Matrix  I  =    

[  3  4    1    1 ]                                               [ 3  ]

   6  6    4    6                                                   5

[  2  2    5    4 ]                                                 2

                                                                      [  6 ]

Matrix  PI  =

[  3*3 + 4*5 + 1*2 + 1*6 ]              [  9 + 20 + 2 + 6   ]              [ 37 ]

   6*3 + 6*5 + 4*2 +  6*6        =       18 + 30 + 8 + 36        =       92 

[  2*3 + 2*5 + 5*2  + 4*6 ]             [  6 + 10 + 10 + 24 ]             [ 50 ]     

Since it is a point of intersection.....the point will satisfy BOTH of the equations....and they are equal at this point...so set them equal to each other

x^2-3x+7  =  5 - x^2      and solve for 'x'     

                                            then use this value of 'x'    (there MAY be TWO values of 'x' !...and there may be NONE !!)  in one of the equations to compute 'y'  

I actually just took this test. 

QTR=55

RQS=47

TS=102

TRS=51

QSR=55

and QT=54 (not pictured)

It is a concave down parabola

the vertex is (2,3)

so

\((x-2)^2=4a(y-2) \qquad where\;\; a<0\)

It appears to pass through (-3,0)

Sub this point in to get the value of a.

Then you will have your equation.

AH!, i see...

Thank you  Heureka, for giving me more understanding on this quesion/problem

Factor the expression given below.
Write each factor as a polynomial in descending order.
\(100w^2 x^2 - 36y^2 z^2\)

\(\begin{array}{|rcll|} \hline && 100w^2 x^2 - 36y^2 z^2 \\ &=& \left(10w x - 6y z\right) \left(10w x + 6y z\right) \\ \hline \end{array}\)

Good story, except one line spoils it:  He looked at the smoking revolver in his hand, and with an oath he hurled it to the roof at his feet. 

I've seen it in a dozen movies.  It's a typical and clichéd portrayed reaction to having used a firearm.  Shooting someone was justified and necessary, but the user feels remorse – or maybe guilt... whatever – and blames the gun for what he himself had done.  What a crock. 

And in addition to the scene being hackneyed and trite, it's out of character.  Earlier, it didn't bother the sniper to gun down an elderly woman who tried to have him killed, nor by the way would it have bothered me. 

.

No. The correct answer is A.  Look at the axes labels.

I think so too 

The graph above is the graph of \(y=x^2\) translated and reflected in the xy-plane.
Find an equation for this graph.

1. translated

\(\begin{array}{|rcll|} \hline y_0 &=& x_0^2 \\ && \boxed{y = y_0+2\\ x=x_0+7 } \\ (y-2) &=& (x-7)^2 \\ y &=& 2 + (x-7)^2 \quad \text{ after translation} \\ \hline \end{array}\)

2. reflected

\(\begin{array}{|rcll|} \hline y_0 &=& 2 + (x_0-7)^2 \\ && \boxed{y = -y_0\\ x=x_0 } \\ -y &=& 2+(x-7)^2 \\ \mathbf{y} &=& \mathbf{-2-(x-7)^2} \quad \text{ after translation and reflextion} \\ \hline \end{array}\)

I think the answer is A because there is one area that is a cluster bundle of dots in one area and one outlier away from all the other points; irrelevant.

I believe the answer is A

i am so tired lmao :') and im bored and i am mad and im sad and i am disappointed and im yeah!!!

well i used to do math competitions like a while back but i became extremely busy so i can't do them anymore but yeah we have like a "math club" thing and its really helpful... i really liked it. They basically get questions from previous math competitions and they use those as practice for the upcoming ones, and whatever was unclear they would explain it to us and yeah :-) 

oh lol :')  

so let me ask you this....
would it have two solutions or-

All reals are also complex numbers.

A complex number need not have an imaginary part.

Friended you all! Thanks!

LOL I went on to compose one of the longest explanations I ever did.

I got as close to you, and when I realized my answer was falling apart, I just deleted the entire thing and waited for someone else to explain XD.

Might help -> https://www.emathhelp.net/calculators/algebra-2/inverse-cosecant-calculator/

Lol my math competitions are starting in the winter.

The kids in my school are highly presitgious in math. At a middle school level, most of us are studying calculus to climb up onto the competition hill. I am considered an amateur at my school lol.

For us, we have a math meeting every friday once the mathclub starts (hasn't started yet)

but the practicing is on us.

The ones who are good practice nearly 5 horus a day.

Like bruh, why does everyone in my school have to be super nerdy...

I don't know because we know that 6 is the mean so then that would leave us with 4 and 5 or 5 and 4 for the mode and the median. 

Yes if we let the mode equal to 4 and the median as 5 then the largest possible value would be 15
however if we let 5 equal to the mode and 4 as the median then wouldn't it be 22?? 

I'm confused because I have two values now, and I don't know what to do. 

Thank you CPhill!! :))