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g(x)  must  be  degree 5  containing the terms  -2x^5  + x^4  +  (some  coefficient)*x^3

So

f(x)  + g(x)  =   (some coefficient ) x^3  + x^2  - 3

1. By similar triangles, EP = 4/7 and FP = 5/3.

The ant can go from the top to the middle, and around, and back to the top.  This gives a maximum distance of five units.

The graph of y = k(x) is the inverse of the graph of j(x), due to the min/max relationship.  So we can invert the graph of the length of the graph of to find the length of the graph of j(x)

Answer: l^2 = 24

#2: V = 1/3*pi*r^2*h

Differentiate both sides to get dV/dt = 1/3*pi*dr/dt*h + 1/3*pi*r^2*dh/dt

Plug in, to get dV/dt = 1.72 inches/second

https://web2.0calc.com/questions/last-question-running-late-please-help

Brilliant Melody.....In all honesty I was getting the same answer (not in such fine style!!) as your solution!

This proves there is an error in the book!! 

Therefore thanks a million and I wish you all the best for the festive season and assure you I am a great fan of yours!!   

Im not too good at probability, but there is a 3/6 chance of it being a prime number (2,3,5) so that would be (3/6)^3 chance of the first 3 being prime, and then a (3/6)^3 for it not to be a prime number. so that would be (3/6)^6

|x + yi| = sqrt(a^2 + b^2)

|aw + b| = sqrt(a^2 w^2 + b^2) = 1

a^2 w^2 = 0, b^2 = 1

a = 0, b = 1 or -1

Two solutions: (a,b) = (0,1) and (a,b) = (0,-1)

https://web2.0calc.com/questions/suppose-that-a-b-and-c-are-positive-integers-satisfying

Construct the perpendicular from E to hypotneuse DF, and apply Pythagoras.  Then DE + DF = 6 + 2*sqrt(3).

Sorry, wrong. thanks for trying!

The degree of g(x) is also 3!

SO the ratio of Allen to Ben is 5:2. First, I would add them up. After that, I divided 1400/7 which equals 200. You have to multiply this by Ben's ratio which is 200*2 and that is 400 sq feet.

160 sq ft.

Median from A = sqrt(2b^2 + 2c^2 - a^2)/2

Median from B = sqrt(2a^2 + 2c^2 - b^2)/2

Median from C = sqrt(2a^2 + 2b^2 - c^2)/2

AB = 24

Plug in to get CG = 3/4*AM = 6.

Plug in the given information. to get (a,b,c) = (4,5,-6).

We can solve parametrically to get x = 5 - 3t, y = t + 2, and z = 2t + 7.  Then x + y + z = 14.

In the coordinate plane, A = (0,0), B = (0,4), and C = (5,0).  If P is a point inside triangle ABC such that AP = sqrt(10) and BP = 3*sqrt(2), then find CP.

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Use the law of cosines to find ∠BAP.    Find ∠PAC

Using ∠PAC and line AP, calculate line segments AD and PD

CP = sqrt[PD² + (AC - √8)²]     ==>   CP ≈ 2.591 

According to this, your answer would be a,b = -1,3

ay = [-5,-4,-3,-2,-1]
be = [1,2,3]
lo = float('inf')
answr = 0
for a in ay:
    for b in be:
        if (1/a+1/b)*(1/b-1/a) < lo:
            lo = (1/a+1/b)*(1/b-1/a)
            answr = [a,b]
print (lo)
print (answr)

Triangle ABC is equilateral with side length 3. A point X is randomly chosen within triangle ABC. What is the probability that X is no more than 2 units away from vertex A?

Hello Guest!

\(\large P=\frac{A_{sec}}{A_{tri}-A_{sec}}\)

\(A_{sec}= \frac{2^2\pi }{6}\\ A_{tri}= \frac{3^2\sqrt{3}}{4}\)

\(P=\large \frac{ \frac{2^2\pi }{6}}{ \frac{3^2\sqrt{3}}{4} - \frac{2^2\pi }{6}} =\frac{\frac{8\pi }{12}}{\frac{9\cdot 3\sqrt{3}}{12}-\frac{4\cdot 2\pi}{12}} =\frac{8\pi}{27\cdot \sqrt{3}-8\pi}\)

\(P= \large 1.162 :1\)

  !

Hello Guest!

\(445:5=89\\ 445=89\times 5\)

  !

Quick review of AoPS Honor Code in case of classes (this is a class question, not alcumus, since when I searched the alcumus forum it yeilded no results.)

Note: I have done more that 5 AoPS classes, so don't tell me that this isn't the case. 

Thank you so much!