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Shown below is rectangle EFGH . Its diagonals meet at Y .
Let X be the foot if an altitude is dropped from E to line FH . If
EX=24 and GY = 25, find the perimeter of rectangle .

$\text{Let $EH$ = x }\\ \text{Let $EF$ = y }\\ \text{Let $EY=HY$ = 25 }\\ \text{Let $HF$ = 50 }$

1.

$\begin{array}{|rcll|} \hline EY^2 &=& EX^2+XY^2 \quad & | \quad XY=HY-H = 25 - HX \\ 25^2 &=& 24^2+(25-HX)^2 \\ 25^2-24^2 &=& (25-HX)^2 \\ \sqrt{25^2-24^2} &=& 25-HX \\ HX &=& 25 - \sqrt{25^2-24^2} \\ HX &=& 25 - 7 \\ \mathbf{ HX} & \mathbf{=} & \mathbf{18} \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline x^2 &=& 24^2 + HX^2 \quad & | \quad HX=18\\ x^2 &=& 24^2 + 18^2 \\ x &=& \sqrt{24^2 + 18^2} \\ \mathbf{ x} & \mathbf{=} & \mathbf{30} \\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline x^2+y^2 &=& HF^2 \quad & | \quad x=30, ~ HF=50 \\ 30^2+y^2 &=& 50^2 \\ y^2 &=& 50^2-30^2 \\ y &=& \sqrt{50^2-30^2} \\ \mathbf{ y} & \mathbf{=} & \mathbf{40} \\ \hline \end{array}$

The perimeter of rectangle:

$\begin{array}{|rcll|} \hline && 2x+2y \\ &=& 2\cdot 30 + 2\cdot 40 \\ &=& 60 + 80 \\ &\mathbf{=}& \mathbf{140} \\ \hline \end{array}$

The perimeter of rectangle is 140

$x+y=a\\ y = a-x\\ x y = x(a-x) =\\ -(x^2 - a x)= \\ -\left(x^2 - ax +\dfrac{a^2}{4}-\dfrac{a^2}{4}\right) = \\ -\left(x-\dfrac a 2\right)^2 + \dfrac{a^2}{4}$

$xy \text{ will be a maximum of }\dfrac{a^2}{2} \\ \text{when }x = \dfrac a 2 \text{ i.e. when}\\ x = y = \dfrac a 2$

$\text{If you draw the force diagram you see that the gravitational force on the rock}\\ \text{can be broken up into a component normal to the inclined ground and a component }\\ \text{along the direction of potential motion}. \text{The magnitude of the component along the direction of potential motion is}\\ |f_m| = 400 \sin(20^\circ) \approx 136.8~ lbs\\ \text{This is the force that must be applied to the rock in the direction of the incline}$

Right triangle ABC has legs AC=21 and BC=28 and hypotenuse AB=35 .
A square is inscribed in triangle ABC as shown below.
Find the side length of the square.

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{AC-x}{x}} & \mathbf{=} & \mathbf{\dfrac{AC}{BC}} \quad & | \quad AC=21, ~BC=28 \\\\ \dfrac{21-x}{x} & = & \dfrac{21}{28} \\\\ \dfrac{21-x}{x} &=& \dfrac{3}{4} \\\\ (21-x)\cdot 4 &=& 3x \\ 21\cdot 4-4x &=& 3x \\ 84&=& 3x+4x \\ 84&=& 7x \\ 12&=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{12} \\ \hline \end{array}$

The side length of the square is 12

What is the greatest number of points of intersection that can occur when
2 different circles and 2 different straight lines are drawn on the same piece of paper?

$\text{Let $n$ the number of circles }\\ \text{Let $m$ the number of straight lines }$

$\begin{array}{|rcll|} \hline n &=& 2\\ m &=& 2 \\\\ && \dfrac{1}{2}\cdot m(m-1)+n(2m+n-1)\\ &=& \dfrac{1}{2}\cdot 2(2-1)+2(2\cdot 2+2-1) \\ &=& 1 +2\cdot 5 \\ &\mathbf{=}& \mathbf{11} \\ \hline \end{array}$

The greatest number of points of intersection is 11

Maybe.

(sqrt(36))^3 = 216

$T(x,y) = x^2 -4x + y^2 + 2y = \\ (x-2)^2 -4 + (y+1)^2 - 1 = \\ (x-2)^2 + (y+1)^2 - 5$

$\text{Clearly the first two terms are non-negative and are zero at }x=2,~y=-1\\ T(2,-1) = -5\\ \text{and this is as cold as it gets}$

Draw square, start at top left corner, name corners clockwise ABCD

Draw diagonal AC as long side of rectangle, complete rectangle counterclockwise ACEG

AD = DC  call each side 1

therefore diagonal AC = sqrt2

Area of rectangle = area of square = 1

Area of rectangle = AC x CE = sqrt2 x CE

Area of rect = area of square = 1

therefore sqrt2 x CE = 1

therefore CE = 1 over sqrt2 or sqrt2 over 2

AC over CE = sqrt2 over (sqrt2 over 2) = 2sqrt2 over sqrt2 = 2 over 1

The answer is A,B,C,D,E

Factor   (x-7)(x-1) = 0      so   x = 1 or 7       the lesser of which is  1   

Rectangle diagonal is 4.

Not enough info or diagram given.....

Suppose the square is 5 x 5     perimeter = 20   area = 25

Rectangle could be  9 x 1         Perimeter = 20   area = 9

Rectangle could be 8x 2           perimeter = 20   area = 16

rectangle could be 4 x 6           perimeter = 20   area = 24

etc.....

Need diagram or more info.....See what I mean?

The total number of perfect squares and perfect cubes for the same number would be: (15!)^(1/6) =104 such numbers. And these numbers will consist of ALL prime numbers up to: 10^[(log(15!))/ log(6)]=~105. The last prime number to be included would be 103.
However, all these prime numbers must be raised to a MINIMUM power of 6 and multiples of 6. So, the first of these numbers would look like this:

2^6,  2^12,  2^18,  2^24,  2^30,  2^36(which is < 15!). Then, they will continue with the next prime: 3^6,  3^12,  3^18....etc. Then you will have a combination of powers such as: 2^6  x  3^6  x  5^6 and multiples thereof.
The computer-generated numbers will begin with:

(1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441, 1000000........etc......and end in: 1000000000000,  1061520150601, 1126162419264, 1194052296529, 1265319018496), for a grand total = 104.

Volume of a cone =   pi  r^2  h/3           r = x/2     h = x

Substitute     pi    (x/2)^2  x/3

                    V = pi x^3 /12           Solve for x

                    12V/pi  =  x^3

                     x = cubrt( 12V/pi)

5040 minutes  =   5040/60 = 84 hours =  84/24= 3 days 12 hours     = Tuesday at 5 AM

   Uh-Huh !

I will assume they did NOT double on Sunday....but rather it took until Monday for them to double to 4

Tuesday would be 8

Wednesday would be 16

Thurs   would be 32

Friday    64

  Saturday 128

That means the answer is y=7/8+6?

This is simple as well...

If there is 4 people and each pay $12...

12*4= $48 

So $48 is the total cost of a video game.

If three people are paying for the game...

48/3=16

But it is asking how much MORE so..

16-12= 4

See previous answer:

https://web2.0calc.com/questions/math-help-pls_4#r1

When I split the cost of a video game equally with 4 friends, we each pay $12.00. How much more do each of us pay if I split the cost equally with only 3 friends?

6:5      11 parts    66/11 = 6 flowers for each part      6 parts roses =   6(6) = 36 roses

Hey Chandler! 

It's simple.

Look,

999,999,999

666,666

333

Are all divisible to 3.

And when added togther the sum is STILL DIVISIBLE BY 3!

So when you add 1,

It is not divisible by three

Then you willl just have a remainder of 1!