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Let m  be the constant amount  (in feet) that it grows every year....so.....at the end of 4 years, the height is

[4  + 4m]

And  at the end of the  6th year the height  is  4 + 6m

And we know that

(1 + 1/5)  [ 4 + 4m ] =  4 + 6m

(1.2) [ 4 + 4m ]  = 4 + 6m

4.8 + 4.8m = 4 + 6m

4.8 - 4   = 6m - 4.8m

.8 = 1.2m        divide both sides by  1.2

.8 / 1.2  =  m  = 8/12  = 2/3 ft  =  8 inches  per year

In rectangle ABCD, AD=15 and point E is on side BC such that EA = 13 and ED = 14.

What is the area of rectangle ABCD?

\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 13^2 \\ \mathbf{x^2} &=& \mathbf{13^2 - y^2} \\ \hline x^2 +(15-y)^2 &=& 14^2 \quad | \quad \mathbf{x^2=13^2 - y^2}\\ 13^2 -y^2 + 15^2-30y+y^2 &=& 14^2 \\ 13^2 + 15^2-30y &=& 14^2 \\ 30y &=& 13^2+15^2 - 14^2 \\ y &=& \dfrac{13^2+15^2 - 14^2}{30} \\ \mathbf{y} &=& \mathbf{6.6} \\ \hline \mathbf{x^2} &=& \mathbf{13^2 - y^2} \\ x^2 &=& 13^2 - 6.6^2 \\ x^2 &=& 125.44 \\ \mathbf{x} &=& \mathbf{11.2} \\ \hline \end{array} \)

Area of rectangle ABCD:

\(\begin{array}{|rcll|} \hline A &=& 15x \\ &=& 15*11.2 \\ \mathbf{A} &=& \mathbf{168} \\ \hline \end{array}\)

15   =  20 e^(k* 10)    divide both sides by   20

3/4  = e^(10k)       taket the Ln of both sides

Ln (3/4) = Ln e^(10 k)    and we can write

Ln (.75)  = (10k) * Ln e        [ Ln e  =1....so....we can ignore this  ]

Ln (.75) /=  10k     divide both sides by 10

Ln (.75)/10  =  k

And we want to know when 10  grams remain....so....

10  = 20e^(t * Ln(.75) / 10 )           divide both sides by 20

.5  = e^(t * Ln (.75)/10 )          using a similar proceedure  as above

Ln .5  =  t * Ln (.75) /  10         multiply both sides by  10 / Ln (.75) and we get that

10 * Ln (.5) /  Ln (.75)  =  t

This is the exact  answer........you will need a calculator from here   [ the answer is ≈  24.09 yrs  ]

Let    AB  =  w

and   BE  =  x

and   EC  =  y

The area of rectangle ABCD   =   ( length )( width )   =   15w

By the Pythagorean theorem, we can make these two equations:

w² + y²  =  14²

w² + x²  =  13²

Since ABCD is a rectangle, BC must be the same length as AD, and so we can make this third equation:

x + y  =  15

Now we can begin solving this system of three equations. Here's one way to do that:

x + y  =  15

y  =  15 - x

w² + x²  =  13²

w²  =  13² - x²

w²  =  169 - x²

w² + y²  =  14²

(169 - x²) + (15 - x)²  =  14²

(169 - x²) + (15 - x)²  =  196

169 - x² + (15 - x)(15 - x)  =  196

169 - x² + 225 - 30x + x²  =  196

-x² + x² - 30x  =  196 - 169 - 225

-30x  =  -198

x  =  6.6

w²  =  169 - x²

w²  =  169 - (6.6)²

w²  =  125.44

w  =  11.2

We don't care what  y  is, we just needed to know what  w  is.

Area of rectangle ABCD   =   15w   =   15(11.2)   =   168   sq units

Find the sum = \(\dfrac{1}{2} + 2*\dfrac{1}{2^2} + 3*\dfrac{1}{2^3} + 4*\dfrac{1}{2^4} + 5*\dfrac{1}{2^5}+ \ldots\)

\(\begin{array}{rcrl} s &=& 1*\dfrac{1}{2} +& 2*\dfrac{1}{2^2} + 3*\dfrac{1}{2^3} + 4*\dfrac{1}{2^4} + 5*\dfrac{1}{2^5}+ \ldots \\ \dfrac{1}{2}s &=& & 1*\dfrac{1}{2^2} + 2*\dfrac{1}{2^3} + 3*\dfrac{1}{2^4} + 4*\dfrac{1}{2^5}+ \ldots \\ \hline s-\dfrac{1}{2}s &=& 1*\dfrac{1}{2} +& 1*\dfrac{1}{2^2}+1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4} +1*\dfrac{1}{2^5}+ \ldots \\ \end{array} \)

\(\begin{array}{rcll} \dfrac{1}{2}s &=& \dfrac{1}{2}* \left( 1+ \dfrac{1}{2^1}+1*\dfrac{1}{2^2} +1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4}+ \ldots \right) \\ s &=& 1+ \dfrac{1}{2^1}+1*\dfrac{1}{2^2} +1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4}+ \ldots \\ s &=& 1+ \dfrac{1}{2^1}+\dfrac{1}{2^2} +\dfrac{1}{2^3} + \dfrac{1}{2^4}+ \ldots \\ \end{array}\)

Now use the formula for the sum of an infinite geometric series.

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{1-\dfrac{1}{2}} \\ s &=& \dfrac{1}{ \dfrac{1}{2}} \\ \mathbf{s} &=& \mathbf{2} \\ \hline \end{array}\)

Two small semicircles are inscribed in a larger semicircle, and a circle is drawn such that it is internally tangent to all three semicircles, as seen above.

If the diameter of the biggest semicircle is 12, find the radius of the small circle.

\(\begin{array}{|rcll|} \hline \mathbf{(6-r)^2 +3^2} &=& \mathbf{(r+3)^2} \\ 36-12r+r^2+9 &=& r^2+6r+9 \quad | \quad -(r^2+9)\\ 36-12r &=& 6r\\ 18r &=& 36 \quad | \quad : 18 \\ \mathbf{r} &=&\mathbf{ 2 } \\ \hline \end{array}\)

If we  connect the centers  ofthe two larger circles.....we can form  the base of  an isoceles triangle that will be  6 units in length

Let the radius of the smaller circle = r

And the two  sides of this triangle will be segments joining the centers of the larger circle with the center of the smaller circle ....and the length of these sides  = radius of larger circles  + radius of smaller circle=  3 + r

And using the  Pythagorean Theorem, we can find the height of this isosceles triangle as

√ [ (3 + r)^2  - 3^2]  =   

√ [ 9 + 6r + r^2 - 9] =

√ [ 6r + r^2 ]

So....the height of this triangle  plus the radius of the smaller  circle =   radius of the larger circle....so...

√[6r + r^2] + r  = 6     subtract  r from both sides

√ [ 6r + r^2]  = r - 6     square both sides

6r + r^2  = r^2 -12r  + 36      

18 r  = 36      divide both sides by 18

r = 2 = radius of smaller circle

ok thanks.

∑[n / 2^n, n, 1, ∞] = 2

See  the following image :

The slope  of  the line containig  side  BD   =     [ 0 - 24 ] / [ -7 - 0]  = 24/7

And the equation of  line containg this segment is given by

y = (24/7)x  + 24        multiply this  by  7

7y  = 24x  +  168     put into  standard form

24x - 7y + 168  = 0       [ Ax  + By  + C  =  0  ]

The circle will be centered at  (0, 0)

And we can use  the formula for the  distance from  point (m, n) = (0,0) to  the line we found = the radius of the  circle

So

l  Am  + Bn  + C l             l  24(0) - 7(0)  + 168  l             168

_______________ =     ___________________   =     _____   =   6.72 =  AF

  √[ A^2 + B^2 ]                  √ [ 24^2 + 7^2 ]                       25

I have programmed my computer with "Modular Multiplicative Inverse" algorithm which calculates them automatically. It is very similar to the algorithm used in this online calculator:

https://planetcalc.com/3311/

If you have hints you should share them with the answerers right froom the very beginning.

Why should answerers have less information than you do ?

 Thanks! 

Let the side of a square  =  S

Then the diagonal of a square  =√2 *S

The diagonal of the rectangle  =  √ [ S^2  + (3S)^2  ]  =   √(10S^2)  =  √10*S

So....the ratio we want is

√2/ √10  =    √ (2/10)  = √ (1/5)  = √1 / √5  =   1  / √5

So

a + b   = 1  + 5    = 6

Q1 -  correct  !!!

Q2  -  we can carry this one step more

arctan  ( 1/ √3)  =  x   =   30°

Good job  !!!

The rectangle is  inscribed in a regular hexagon

(1/2)  of the width of the rectangle =  √3/2

So  the width is 2 times this  =√3  units

The height ,H, of the rectangle can be found as   2  +  2 (1/2)   =  3 units

So.....the area  of the rectangle =  H * W   =  3√3  units^2

Could you please explain how you got those numbers?

Thanks guest.

Here is another attack method:

\(log_2\;\frac{8}{\sqrt{32}}\\~\\ =log_2\;\frac{2^3}{({2^5})^{0.5}}\\~\\ =log_2\;\frac{2^3}{2^{2.5}}\\~\\ =log_2\;(2^{3-2.5})\\~\\ =log_2\;(2^{0.5})\\~\\ =0.5\;log_2\;(2)\\~\\ =0.5\;*1\\~\\ =0.5\)

It states in that a and b are real.  The coefficients or z are clearly not real.

Thanks EP.  

I want to give it a go myself.   

z^2 + (a + 8i) z + (-39 + bi) = 0.

\([z-(4-5i)][z-(p+qi)]=z^2 + (a + 8i) z + (-39 + bi) \\ \text{Equating coefficients}\\ -(4-5i)-(p+qi)=a+8i\qquad and \qquad (4-5i)(p+qi)=-39+bi\\ -4+5i-p-qi=a+8i\qquad \qquad and \qquad 4p+5q+4qi-5pi=-39+bi\\ -4-p+(5-q)i=a+8i\qquad \qquad and \qquad 4p+5q+(4q-5p)i=-39+bi\\ -4-p=a \qquad 5-q=8\qquad \qquad and \qquad 4p+5q=-39 \qquad 4q-5p=b \\ -4-p=a \qquad \color{red}q=-3 \color{black}\qquad \qquad and \qquad 4p+5q=-39 \qquad 4q-5p=b \\ -4-p=a \qquad \qquad \qquad \qquad and \qquad 4p-15=-39 \qquad -12-5p=b \\ -4-p=a \qquad \qquad \qquad \qquad and \qquad \qquad \color{red}p=-6 \color{black}\qquad -12-5p=b \\ -4+6=a \qquad \qquad \qquad \qquad and \qquad \qquad \qquad \quad \color{black}\qquad -12+30=b \\ \color{red}a=2 \qquad \qquad\qquad \qquad \qquad and \qquad \qquad \qquad \quad \qquad\qquad b=18 \\\)

So we have

\([z-(4-5i)][z-(-6-3i)]=z^2 + (2 + 8i) z + (-39 + 18i) \\\)

I have not rechecked by expanding.     But I also get the other root to be -6-3i.

Is there an easier way to do it?

Coding:

[z-(4-5i)][z-(p+qi)]=z^2 + (a + 8i) z + (-39 + bi) \\
\text{Equating coefficients}\\

-(4-5i)-(p+qi)=a+8i\qquad and  \qquad (4-5i)(p+qi)=-39+bi\\
-4+5i-p-qi=a+8i\qquad \qquad and  \qquad 4p+5q+4qi-5pi=-39+bi\\

-4-p+(5-q)i=a+8i\qquad \qquad and  \qquad 4p+5q+(4q-5p)i=-39+bi\\
-4-p=a \qquad 5-q=8\qquad \qquad and  \qquad 4p+5q=-39 \qquad  4q-5p=b \\
-4-p=a \qquad \color{red}q=-3 \color{black}\qquad \qquad and  \qquad 4p+5q=-39 \qquad  4q-5p=b \\

-4-p=a \qquad \qquad \qquad \qquad and  \qquad 4p-15=-39 \qquad  -12-5p=b \\
-4-p=a \qquad \qquad \qquad \qquad and  \qquad \qquad \color{red}p=-6 \color{black}\qquad  -12-5p=b \\
-4+6=a \qquad \qquad \qquad \qquad and  \qquad \qquad \qquad \quad \color{black}\qquad  -12+30=b \\
\color{red}a=2 \qquad \qquad\qquad \qquad \qquad and  \qquad \qquad \qquad \quad \qquad\qquad  b=18 \\

Let the  longer side of the  small rectangle  = x

Let the shorter side   =  y

For the width of the square we have  that

x + ( x - y)  + x  + (x - y) = 32

4x - 2y  = 32

2x - y  = 16  ⇒  -2x + y = -16    (1)

For the  height of the square we have that

y  + x  +  y  + x + y  = 32

2x + 3y  = 32   (2)

Add  (1) and (2)  and we have that

4y  = 16

y = 4

And x = 10

We can decompose the yellow area  into  2 rectangles

One has a width  of   (32 - x)  and a height of  (x+ y)

So...its area  =  (32 - 10) ( 10 + 4)  = (22) (14)  =  308

And the other rectangle has a width of (x - y)  and a height of (x + y)

So.....its area  =  (10 - 4) ( 10 + 4) =  (6)(14)  =  84

So....the yellow area =  308 + 84  =   392 units^2

Sum of the inverses of the first 12 positive integers:

1 + 7 + 9 + 10 + 8 + 11 + 2 + 5 + 3 + 4 + 6 + 12 =78

78 mod 13 =0

Here is the Web2.0 Calculator version:

sin (40) sin (80) + sin(80) sin (160) + sin (160) sin (320) = 0.7500000000002045   ~~3/4