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WE CAN WRITE THE EQUATION LIKE \(\begin{pmatrix} 1&&2&&1\\ 2&&-1&&2\\ 3&&1&&-1 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} 5\\ 15\\ 8 \end{pmatrix}\)

SOLVING THE EQUATIONS WE GOT X=4, Y=-1 , Z = 3

so, our answer is \(\begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix}\).

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The image is in asymptote... please don't answer as it is most likely from an AoPS course.

1

  C

                4

                       X

                                 12

B                                           A

triangle  XBA similar to triangle BCA

triangle  XCB similar to triangle BCA

So

triangle XCB is similar to triangle XBA

So

CX / XB  = XB / AX

4 / XB   XB / 12

XB^2  =  48

XB  = sqrt (48)  =  4 sqrt (3)

2

[ABC]  =  (1/2) (AB) ( BC) sin (ABC)   =

(1/2) ( 6) (8)  (sqrt 2) / 2  =

12 sqrt 2

PLEASE SEND THE DIAGRAMS TO SOLVE THE PROBLEM 

PLEASE SEND THE DIAGRAM TO SOLVE THE QUESTION.

answer is 12\(\sqrt2\).

explain :

DRAW A LINE X-B-C AND CX PERPENDICULAR TO AX .

\( \angle\)ABC=135 .THEN \( \angle\)ABX=45=\(\angle\)BAX.(\(\angle\)AXB=90)

45:45:90=1:1:\(\sqrt2\)

AB=6 ,SO AX=BX=6/\(\sqrt2\)=3\(\sqrt2\).

BC=8.

AREA  OF TRIANGLE ABC=(1/2)*AX*BC=(1/2)*3\(\sqrt2\)*8=12\(\sqrt2\).

This is simple geometry. The triangles ABC,BXC,AXB are all similar. You are given AX=12,XC=4=12. So it is easy to find the other lengths. Eg XC/BC=BC/AC, so BC=8 and hence BX=4sqrt3.

Do you have developed an AI model that can do that?

Question 1

Six children are each offered a single scoop of any of 3 flavors of ice cream from the Combinatorial Creamery. In how many ways can each child choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children?

We can select 3 children from any  of the  6  to have the same  flavor  =  C (6,3) =  20  different sets  of  children

And each of these sets  has  three ways to  choose the same  flavor  = C (3, 1) = 3

The remaining 3 children are by default.....one  of then  can choose any  of  the  2  remaining flavors and  this  child can be any one  of the remaining  3...the other 2 children are stuck with the  remaining  flavor not yet selected...so the  total sets here  =  C(3,1) * C(2,1)

So....the total  number of ways = 

C(6,3) * C (3,1) * C(3,1) * C(2,1)  =

20  * 3  *  3 * 2  = 

360  ways

EXPLAIN:

MP=QM=18.

DRAW AY PERPENDICULAR TO PR ,THEN,

IN TRIANGLE PYM AND PYA

PY IS COMMON SIDE .

ANGLE MPY = YPA

ANGLE PMY=PAY

SO, PYM=PYA

SO, TRIANGLE PYM AND PYA ARE CONGRUENT.

MY=AY=8,

SO AREA OF PYR=(1/2)*22*8=88.

ANSWER IS 88.

answer is 88 degree.

explain :

AIB ANGLE =134 DEG

IAB ANGLE  + IBA  ANGLE =180-134= 46 DEG.

BY AND AX ARE BISECTORS OF ANGLE CAB AND CBA.SO,CAB+CBA=46*2=92 DEG.

ACB OR C ANGLE =180-92=88 DEGREE.

The following is one possibility:

n choose 2 is odd when n=2,3,6,7,10,11,14,15,18,19,22,23,26,27,30,31,34,35,38,39,42,43,46,47,50,51,54,55,58,59,62,63,66,67,70,71,74,75,78,79,82,83,86,86,87,90,91,94,95,98,,99

total=50 ways.

n choose 2 is even when n=4,5,8,9,12,13,16,17,20,21,24,25,28,29,32,33,36,37,40,41,44,45,48,49,52,53,55,56,57,60,61,64,65,68,69,72,73,76,77,80,81,84,85,88,89,92,93,96,97,100.

https://web2.0calc.com/questions/for-how-many-integers-n-where-2-n-100-is-n2-odd 

answer is wrong.

a. That's not enough to say. What are a, b, and c? 

Also, you'll want to learn how to type squares. One way is to write a^2 + b^2 = c^2.

e. Not quite. The units are wrong.

a)

There are 12 people and we want to pick 4 of them. So, we use

 \(\binom{12}{4} = 495\)

b)

First, we know that Mark has to be on the committee, so we put him in immediately. This leaves 3 spots to be filled by 11 people. Hence we use

\(\binom{11}{3} = 165\)

c)

We know that mark cannot be included, so there are 11 people who can go in 4 spots.

\(\binom{11}{4} = 330\)

I hope these are correct!