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Looking at the AoPS Honor Code here: https://artofproblemsolving.com/school/handbook/current/honorcode which seems to pop up every class that starts for you, it clearly says that "You should not search for potential solutions to problems online or elsewhere. (Especially searching for the problem statement!) Using the internet in this way to search for answers is against our Course Policies! Instead, if you need some assistance with a particular problem, we welcome you to ask questions on your course message board. Your instructor and other AoPS staff members will be happy to help. Again, the solution that you submit should be your own solution in your own words. If you copy a solution from the internet into your submission, this is misrepresenting another person's work as your own. If you are reading a solution from the internet as you create your submission, this is misrepresenting another person's work as your own." So no.

If it is simple, why don't you do it yourself?

Rather than asking us that question,  why don't you just honour the agreement you have made with your online school.

If you have agreed to not ask for external help and then you ask anyway,  You are cheating.

Do you think we should encourage blatant cheating on this site?

Find all solutions to the system
a + b = 14
a^3 + b^3 = 812 + 3ab

\(\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a^3 + b^3 = 812 + 3ab,~ a+b=14\\ 14^3 &=& 812 + 3ab+3*14*ab \\ \ldots \\ ab &=& \dfrac{1932}{45} \quad | \quad b = 14-a \\ a(14-a) &=& \dfrac{1932}{45}\\ 14a-a^2 &=& \dfrac{1932}{45}\\ \mathbf{a^2-14a+\dfrac{1932}{45} } &=& \mathbf{0} \\ a_1 &=& 9.4630604269 \\ a_2 &=& 4.5369395731\\\\ \qquad \mathbf{b} &=& \mathbf{14-a} \\ b_1 &=& x_2 \\ b_2 &=& x_1 \\ \hline \end{array}\)

Troll solution: https://www.desmos.com/calculator/fyaevo8phf shown on the graph, asymptote is x=2 . Y infinitely goes up and when y is infinitely up x will be undefined

Thus when y = busy beaver ^ busy beaver … (busy beaver amount of busy beavers in the exponent which is exponented by) x will be undefined.

Dunno about vectors, but in trig we have that sin^2(theta) + cos^2(theta)=1 which is like the pythog theorem

Hello, LaTeX is rendered as an image and please make sure to type those fancy numbers out next time.

Question: 

A school offers both Spanish and Chinese language classes. This past week, 10 students dropped from the Chinese classes, which increased the number of students who take neither class from 27 to 31. How did this change impact the number of students who take the Spanish class but not the Chinese class?

A school offers both Spanish and Chinese language classes. This past week, 10 students dropped from the Chinese classes, which increased the number of students who take neither class from 27 to 31. How did this change impact the number of students who take the Spanish class but not the Chinese class?

31-27=4 people totally dropped out of both

10-4 = 6 people more who have dropped out of Chinese but still take Spanish.

Top Left.

https://media.officedepot.com/image/upload/b_rgb:FFFFFF,c_pad,dpr_1.0,f_auto,h_1665,q_auto,w_1250/c_pad,h_1665,w_1250/v1/products/369979/369979_o51?pgw=1&pgwact=1

These people... >:( 

Please actually ask a question that makes sense. 

"...increased the number of students who take neither class from  to ." 

From what to what? 

Volume of 1 'batch' =   9 * 13 * 1 = 117 in³

   volume of 1.5 batch   =   1.5 * 117 = 175.5  in³

12 * 15  * d   = 175.5

d = depth = .975 inch ~~  .98 inches

(2z+4)*2       +       16z+12 = 180       (inscribed angles intercept arcs twice their measure)

z =  8

Arc BA = 16z+12 = 140 degrees

CA and DB are diameters of the circle and EC, ED, EB, and EA are radii which means EC=ED=EB=EA which means they are isosceles triangles and since angle CED is congruent to angle AEB we have triangle AEB is congruent to triangle CED.

Now we need ti find z.

Let's focus on triangle CED. Since triangle CED is an isosceles triangle, angle C and D are congruent which means

16z+12+2z+4+2z+4 = 180

20z+20=180

20z=160

z=8

Then solve for angle D

2z+4 = D

16+4 = D

D = 20

And since triangles CED and AEB are congruent Angle D = Angle A so Angle A = 20

:)))))))))))))))))))))

Wait oops you asked for mBA

If thats the case it is quite simple. We already know z = 8 and angle CED = angle AEB and since CED is 140 that means AEB is 140

SO IN CONCLUSION

mBA = 140

Wow i wrote alote

Hint:   AB = BC = 12

Area of black and white square = 2/3(16)

2/3(16) = 10 ²/₃

Distance = sqrt(2 * 10 ²/₃)

l x w x h = 8

l = w = h = 2

side length = 2 ft

2x2 = 4 ft²   = surface area of 1 side    there are 6 sides     = 24 ft2

Vertex 1,-3

Parabola vertex form     y = a (x-1)^2 - 3      sub in one of the points (3,2) to calculate 'a'

                                       2 = a (3-1)^2 - 3

                                       2 = 4a - 3

                                         a = 5/4

y = 5/4 (x-1)^2 -3     is your parabola    expand it to get the 'quadratic'

f(x) = 5/4 x^2 - 5/2 x + 5/ 4  - 3

      = 5/4 x^2 - 5/2 x - 7/4      

The possible values of x are 20, -20, 35, and -35.

The path is an ellipse centered at (1,0), with the semi-minor axis (in the x-direction) of 1, and the semi-major axis (in the y-direction) of 3.

x,y,z are positive integers. find all triples that work

4,cw

A 12''-diameter pizza and a 16''-diameter pizza are each cut into eight congruent slices. Jane ate three slices of the 12'' pizza. Mark ate four slices of the 16'' pizza. How many more square inches of pizza did Mark eat than Jane? Express your answer as a common fraction in terms of pi.  

Area of the 12" pizza  =  πr²  =  π • 6²  =  36π   Jane ate 3/8 of this, which would be 13.5π  

Area of the 16" pizza  =  πr²  =  π • 8²  =  64π   Mark ate 4/8 of this, which would be 32π  

Mark ate 32π – 13.5π  =  18.5π more square inches than Jane ate  

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