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-15 is the constant :/

Thanks Heureka :)

Ok Ginger.

I do not know enough about this to argue with you.

I suppose I should do some homework :)

3.

Since DE  = 2EC....then

EC + 2EC  = DC

3EC  = 20

EC  = 20/3

And DE  = 2EC  = 2(20/3)  = 40/3

Note that triangle  AFB  is similar to  triangle EFD

So

AB/ED  = FB / FD

So

20/(40/3)  = FB / FD

60/40 =  FB / FD

6/4  = FB /FD

3/2 = FB / FD

Then BF  consists of 5 equal parts.....FB is 3  of them and FD  is 2 of  them

And in triangle BDC, FG  is drawn parallel to DC....and we have the following relationship :

BF / BD  = FG / DC

3 / 5   = FG / 20      multiply both sides by 20

20 * 3  / 5    = FG

60 / 5  = FG

12  = FG

Hey, I think you made a little mistake at the begining...

I think the equation should be (2+i)-(-4)+(-i)+(2+4i)...

2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.

The area  of the triangle will be  (1/2)(10) (24)  = 120  units^2

So....if we let  10  be one altitude....the shortest altitude  will be  drawn  from the intersection of the legs  to the hypotenuse

And since the areas  are equal....we have that

(1/2) (length of the shortest altitude) (26)  = 120    

(13) (length of shortest  altitude)  = 120      divide both sides  by 13

length of the shortest altitude  = 120 / 13   units  .....just as CluelesssPersonn  found  !!!!

Yeah...  Why's that? I don't know, but I am guessing that is because somebody answer it and it was long so after a while if no more people come to answer it, it locks it? AND maybe nobody come back and view the question...

And maybe because the one who anwer the question edited it, after I published this and edit it once again. I noticed it became locked!! 

The original D  is  ( 4,1)

When reflected across the x axis, 'D'   is  ( 4, -1)

D' and  D" will lie on a line that  will be  perpendicular to  y  = x + 1

So....the slope of this line is  - 1

And the equation of this line   is

y  = - ( x - 4) - 1

y = -x + 4 - 1

y = -x + 3

The  intersection  of these two lines  will be the midpoint  of   D'  and D"

So...setting the y's equal, we have

-x + 3  = x + 1       add  x to both sides,  subtract 1 from each side

2   = 2x     divide both sides by 2

1  = x

And using either line to  find the y coordinate of this intersection point, we have that

y = x + 1

y = 1 + 1

y  = 2

So...the midpoint  of  D' and D"  is  ( 1, 2 )

So...using the midpoint formula with the points ( 4, -1) and (1,2), we can  find  D'  as

(x + 4)/2  =   1                                ( y + -1) / 2  = 2

Multiply  both sides  by 2

x + 4  = 2                                           y - 1  = 4

subtract 4 from both sides                 add 1 to each side

x = -2                                                 y = 5

So...  D"    =  (-2, 5)

Here's a graph that shows this :

https://www.desmos.com/calculator/rbi6tjxihp

We need to  know what the coordinates  of the original  "A" are  ....  

The protractor is considered to be  "cheating"  by Euclidians....!!!!!

I need the explanation to the same problem.

Can you please help, I need it fast

How many ways are there up an 11 step staircase if you can go one or two steps at a time?

Let \(F_n\) be the number of ways to climb n stairs taking only 1 or 2 steps.

1. Solution:

\(\small{ \begin{array}{|rcll|} \hline F_n &=& ^{n}C_0 +\ ^{(n-1)}C_1+\ ^{(n-2)}C_2+\ ^{(n-3)}C_3+\ ^{(n-4)}C_4+\ ^{(n-5)}C_5 + \ldots +\ ^{(n-k)}C_k\,\quad (n-k) \ge k \\\\ && \text{case of n=11 steps}: \\\\ F_{11} &=& ^{11}C_0 +\ ^{10}C_1+\ ^{9}C_2+\ ^{8}C_3+\ ^{7}C_4+\ ^{6}C_5 \\ &=& 1 +10+36+56+35+6 \\ &=& 144 \\ \hline \end{array} } \)

2. Solution:

\(\begin{array}{|rcll|} \hline F_n &=& \mathcal{F}_{n+1} \qquad \mathcal{F} \text{ is the Fibonacci number }\\\\ && \text{case of n=11 steps}: \\\\ F_{11} &=& \mathcal{F}_{12} \qquad \\ && \mathcal{F}_0 = 0 \\ && \mathcal{F}_1 = 1\\ && \mathcal{F}_2 = 1\\ && \mathcal{F}_3 = 2\\ && \mathcal{F}_4 = 3\\ && \mathcal{F}_5 = 4\\ && \mathcal{F}_6 = 5\\ && \mathcal{F}_7 = 13\\ && \mathcal{F}_8 = 21\\ && \mathcal{F}_9 = 34\\ && \mathcal{F}_{10} = 55\\ && \mathcal{F}_{11} = 89\\ && \mathcal{F}_{12} = 144\\ && \ldots \\ F_{11} &=& 144 \\ \hline \end{array}\)

Thanks!

Assuming x, y and z are real positive numbers satisfying:

\(\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*}\)
\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*}
then, what is the value of xyz?

\(\begin{array}{|lrcll|} \hline & xy-z&=& 15 \\ (1) & \mathbf{xy} &\mathbf{=}& \mathbf{15 + z} \\\\ & xz-y&=& 0 \\ (2) & \mathbf{xz} & \mathbf{=}& \mathbf{y} \\\\ & yz-x&=0 \\ (3) & \mathbf{yz} & \mathbf{=}& \mathbf{x} \\\\ \hline (2)+(3): & xz + yz &=& y + x \\ & z(x+y) &=& x+y \\ & z &=& \dfrac{x+y}{x+y} \\ & \mathbf{z} & \mathbf{=}& \mathbf{1} \\\\ \hline & xy & = & 15 + z \quad & | \quad z = 1 \\ & xy & = & 15 + 1 \\ & \mathbf{xy} &\mathbf{=}& \mathbf{16} \\\\ & xyz & = & xy\cdot z \quad & | \quad xy = 16 \qquad z = 1 \\ & xyz & = & 16\cdot 1 \\ & \mathbf{xyz} &\mathbf{=}& \mathbf{16} \\ \hline \end{array}\)

What is the smallest real number $x$ in the domain of the function $$g(x) = \sqrt{(x-3)^2-(x-8)^2}~?$$

\(\begin{array}{|rcll|} \hline (x-3)^2+(x-8)^2 &=& 0 \\ x^2-6x+9-x^2+16x-64 &=& 0 \\ 10x +9 - 64 &=& 0 \\ 10x - 55 &=& 0 \\ 10x &=& 55 \\ \mathbf{x}& \mathbf{=} & \mathbf{5.5} \\ \hline \end{array}\)

The smallest real number x in the domain of the function is 5.5

4 is wrong

1: 4√3

2: 120/13

3: 12

4: 6*(√3)/3

5: 60/37

6: 216 + 72√3

PS.  I am not sure about all of these answers

Hannah has 2000 feet of fence. She is going to make a rectangular garden using the fence and her barn for one side (so she only needs fencing on three sides of the garden). What is the largest area she can enclose? [Answer with just the number of square feet of area].

Let  two  equal sides of the fence  =  S

Then the remaining side  (opposite the barn)   is   just  2000  - 2S

So....the area, A,   is just

A  = S (2000 - 2S)    simplify

A  = 2000S  - 2S^2         take the derivative of this and set to 0

A'  = 2000 - 4S  = 0

Solving, we have

2000  - 4S  = 0      add 4S  to both sides

2000  = 4S   divide both sides by 4

500  = S

So....the max area  is  S  (2000 - 2S)   =      500 ( 2000 - 2*500)   =   500 (1000)  =   500,000  ft^2

Note that this just boils down to counting the arrangements  of  sets

To get from A to the intermediate point  we can go

(East, East, North, North, North)

And  the total number of arrangements of this set  is  just place East  in any of the two of the  five positions  (or, alternatively, North in any of three of the five positions)...so  C(5,2)   = C (5,3)   = 10  ways

Next...to get from the intermediate point to B we have

(East, East, East , East, North , North , North)

So....the possible arangements of this set  is either placing East  in any four of the seven positions or to place North in any of three of the seven positions  =   C( 7,4)  =  C(7,3)   =   35  ways

So...the number of ways to  get from A to B  =   10 * 35   = 350  ways

Here's  4

ax^2   + 8x  +  4   = 0 

If this has only one solution.....then it will be true  that the  discriminant given by  (8)^2  - 4 (a) (4)  will = 0

So we have

(8)^2  - 4 (a) (4)   =  0   simplify

64  - 16a    = 0      add   16a  to both sides

64   = 16a        divide both sides by 16

4  =  a 

Heureka didn’t make a correction because nothing is incorrect about the list. The question asks for the 17^th alphabetical permutation –that’s all that’s needed for a correct answer. Heureka alphabetized the list because it bugs you; he’s a nice person. I’m not; I would have deranged all of them except for #17, just to make a point. 

A curious question: When you eat alphabet soup, do you alphabetize the spoonful of letters before you eat it?

You might like one of these more.

Words alphabetized:

ehnW ouy aet aabehlpt opus, do ouy iaabeehlptz eht flnoopsu of eelrstt beefor ouy aet it?

Word order alphabetized:

alphabet alphabetize before do eat eat it? letters of soup, spoonful the When you you you

Both words and word order alphabetized:

aabehlpt aet aet beefor do eelrstt ehnW eht flnoopsu iaabeehlptz it? of opus, ouy ouy ouy

You now might have some interesting dictionary dreams. BTW, a dictionary is a book used in the old days to look up the spelling and meaning of words.

GA

Here's  (1)   without a calculator....note that the numerator and  denomonator  are  both just the difference of squares...so we have :

(2011 + 2006) ( 2011  - 2006)       

_______________________    =

(2010 + 2007) ( 2010 - 2007)

(4017) (5)

_________  =

(4017) (3)

 5

__

 3

a)

x + y  = -1

3x  = 4  - 3y

Rearrange the first equation as    y  = -1  - x

Sub this into  the second equation for  y

3x  = 4  - 3(-1 - x)   simplify

3x = 4 + 3 + 3x

3x  = 7 + 3x       dubtract   3x from both sides

0  = 7      this is absurd....so....this system has no solution

b)

3x - 4y  + 2   = 0  ⇒     3x  -  4y  = -2       (1)

10 - 10y  = 10y - 15x

Divide the second equation through by   5 and we have

2 - 2y  = 2y   - 3x       rearrange as

3x - 4y  = -2       (2)

Notice that equations 1  and 2 are exactly the same.....so....when this happens....we have  infinite solutions for both  x and  y