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Differentiate the expression with respect to x.  Set the result to zero and find the value of x which makes it zero.  Plug this value back in to the first expression to find the maximum value of the expression.

Ooh. Need to let Alan the Grapher do this one.

I'm not experienced enough!

I'll delete my answer now.

Melody, you actually do know how to do this….

I'd like to know how to do this too.

I would like you to expand on that answer please.

I think you have copied that answer and have no understanding of how to do it.

AB is a line segment and M is its midpoint.
Semicircles are drawn with AB , AM and MB as diameters on the same side of line AB.
A circle with center O is drawn which touches the three semicircles,
then the radius of the smallest circle is equal to \(\dfrac{x}{y}*AB\).
Find \(\dfrac{x}{y}\).

\(\text{Let the radius of the smallest circle $=r$ }\\ \text{Let $R=\dfrac{AB}{4}$ }\)

\(\begin{array}{|rcll|} \hline (\mathbf{R+r)^2} &=& \mathbf{R^2+(2R-r)^2} \\ R^2+2Rr+r^2 &=& R^2+ 4R^2-4Rr+r^2 \\ 2Rr &=& 4R^2-4Rr \\ 2Rr+4Rr &=& 4R^2 \\ 6Rr &=& 4R^2 \quad | \quad :6R \\\\ r &=& \dfrac{4}{6}*R \\\\ r &=& \dfrac{2}{3}*R \quad | \quad R=\dfrac{AB}{4} \\\\ r &=& \dfrac{2}{3}*\dfrac{AB}{4} \\\\ r &=& \dfrac{1}{3}*\dfrac{AB}{2} \\\\ \mathbf{r} &=& \mathbf{\dfrac{1}{6}*AB} \\ \hline \end{array}\)

\(\mathbf{\dfrac{x}{y}=\dfrac{1}{6}} \)

If the parabola has the x value of the vertex at 5, and a point on the x axis at x = 2  (which is 5-3), then it must also have a point on the x axis at (5+3,0) or at (8,0).

So you have the following two equations;

0 = a*2² + b*2 +c

0 = a*8² + b*8 + c

The slope at the vertex is zero, and the slope is given by dy/dx = 2a*x + b, so x = -b/2a.  We know this x is 5 when the slope is zero, so:

5 = -b/2a

You now have three equations to find the three unknowns, a, b and c.  

I'll leave the rest to you.

Hmm!

" 1/2 of the area of each side is below the midline "

The pyramid is cut horizontally, NOT vertically!!!

Bottom square      A_b = 400

Top square            A_t = 100

Sides                     A_s = 600

Total area              A = 1100 units squared  

N mod  2 = 1

N mod 3 = 2

N mod 4 = 3

N mod 5 =4

N mod 6 =5

N mod 7 =0

Using Chinese Remainder Theorem plus Modular Multiplicative Inverse which are incorporated in this short computer code, we get the following:

i=0;j=0;m=0;t=0;a=(2, 3, 4, 5, 6, 7);r= (1, 2, 3, 4, 5, 0);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

OUTPUT =420n  +  119, where n =0,  1, 2, 3........etc.

Therefore, the smallest number of eggs he had was = 119 eggs.

GCF(LCM(8, 14), LCM(7, 12))

GCF[56,  84] = 28

Nope, it cannot be 80!!! 

A = ( 5² + 4² + 3² ) - 1/2 ( 5 * 12 ) = 20 

Find the minimum value of the product \(P(x,y,z)=(2x+3y)(x+3z)(y+2z)\), when \(xyz=1\) and \(x\), \(y\), \(z\) are positive real numbers.

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{2x+3y}{2} & \ge & \sqrt{2x3y} \\\\ \dfrac{x+3z}{2} & \ge & \sqrt{x3z} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y2z} \\\\ \hline \left( \dfrac{2x+3y}{2} \right) \left( \dfrac{x+3z}{2} \right) \left( \dfrac{y+2z}{2} \right) & \ge & \sqrt{6xy}\sqrt{3xz} \sqrt{2yz} \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & \sqrt{36x^2y^2z^2} \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6xyz \quad | \quad xyz=1 \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6*1 \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6 \\\\ (2x+3y)(x+3z)(y+2z) & \ge & 6 *8 \\\\ (2x+3y)(x+3z)(y+2z) & \ge & 48 \\\\ \mathbf{ 48 } & \le & \mathbf{(2x+3y)(x+3z)(y+2z)} \\ \hline \end{array}\)

The minimum value is \(\mathbf{ 48 }\)

ABC has a right angle at B. \(AC = 60\), \(AE = 52\) and \(CD=39\).
Find the length of \(DE\).

\(\text{Let $AD=t$ } \\ \text{Let $BD=w$ } \\ \text{Let $BE=u$ } \\ \text{Let $CE=v$ } \\ \text{Let $DE=\color{red}x$ }\)

\(\begin{array}{|lrcll|} \hline (1): & 60^2 &=& (u+v)^2+(t+w)^2 \\ (2): & 52^2 &=& u^2+(t+w)^2 \\ (3): & 39^2 &=& (u+v)^2+w^2 \\ (4): & x^2 &=& u^2+w^2 \\ \hline (4)+(1)-(2)-(3): \\ &x^2+60^2-52^2-39^2 &=& u^2+w^2 \\ &&& +(u+v)^2+(t+w)^2 \\ &&& - \left(u^2+(t+w)^2\right) \\ &&& -\left((u+v)^2+w^2 \right) \\\\ &x^2+60^2-52^2-39^2 &=& u^2+w^2 \\ &&& +(u+v)^2+(t+w)^2 \\ &&& - u^2-(t+w)^2 \\ &&& -(u+v)^2-w^2 \\\\ &x^2+60^2-52^2-39^2 &=& 0 \\ &x^2 &=& 52^2+39^2-60^2 \\ &x^2 &=& 625 \\ &\mathbf{x} &=& \mathbf{25} \\ \hline \end{array}\)

The length of DE is \(\mathbf{25}\)

  r = sqrt( 2 * 5² ) = 7.071067718

  A  = r² pi =  50*pi = 157.0796327 units squared  

a = -9, b = 14,

or a = 14, b = -9,

but either way, a+b= 5

Area is   50*pi = 157.0796327 

FTFY:

What is the slope of the line determined by any two solutions to the equation \( \frac{2}{x}+\frac{3}{y} = 0\) ?

Express your answer as a common fraction.

If \(x^2 = 16\), then the possible values of x are \(\pm 4\), meaning 4 and -4.

\(4+-4=0\), so the sum of all possible values of x are 0. 

Angle   MQS = 55 degrees

A circle has a chord of length 10, and the distance from the center of the circle to the chord is 5. What is the area of the circle?

Hello Guest!

\(r^2=(\frac{chord}{2})^2+(dist.)^2\\ r=\sqrt{(\frac{5}{2})^2+5^2}\)

\(r=\sqrt{\frac{25+100}{4}}=\sqrt{\frac{125}{4}}\)

\(A=\pi r^2=\pi \cdot \frac{125}{4}\)

\(A=98.175\)

\(The\ area\ of\ the\ circle\ is\ 98.175\ .\)

  !