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The diagonals bisect each other.....therefore  EY = GY = 26  and EG =  52  = HF

And because  triangle   EYX  is right then    XY  = sqrt ( EY^2 - EX^2) =  sqrt (26^2 - 24^2) = sqrt 100   = 10 

But the diagonals of  a  rectangle are congruent  so   EY = FY = GY =  HY

And   HX = HY - XY  =  26   - 10    =   16

And since triangle EHX is right, then   EH   = sqrt ( HX^2 + EX^2) =  sqrt ( 16^2 + 24^2 ) =   sqrt ( 832)

And triangle HEF is also right so EF  =sqrt  ( HF^2 - EH^2 )  =  sqrt (  52^2 - 832)  = sqrt (1872)

So.....the perimeter  of   EFGH  = 2 [ EH + EF ]  =    2 [ sqrt 832  + sqrt 1872 ]   ≈  144.22

End of year:

   Principal       Interest Rate      Interest earned

1 - $800               8%                      $64

2 - $1,600            8%                      $128

3 - $2,400            8%                      $192

4 - $3,200            8%                      $256

5 - $4,000            8%                       $320

6 - $4,800            8%                       $384

7 - $5,600            8%                       $448

8 - $6,400            8%                       $512

9 - $7,200            8%                        $576

10-$8,000            8%                        $640

Principal + Interest earned over a period of yeas at 8% simple interest=$8,000 + $640==$8,640

P.S. This is exactly the same as investing $8,000 at 8% simple interest for 1 full year.

Let's see:

100 = 2 x 2 x 25  or 4 x 5 x 5 so it works

101 is prime so it is excluded

102 = 2 x 3 x 17 so it works

103 is prime so it is excluded

104 = 2 x 2 x 26 or 2 x 4 x 13 so it works

105 = 3 x 5 x 7 so it works

106 = 2 x 53 ============ this can't be encoded; so the answer is 106

Division Prop of Equality   (we would divide both sides by  -4  to solve for  x )

Since AB = 3 and  BC = 4, by the Pythagorean Theorem, AC = sqrt [ AB^2 + BC^2]  = sqrt [ 9 + 16] =

sqrt (25)  = 5

And since AE = 6 and AC = 5, the area of  quadrilateral  ACGE =   6 * 5 = 30

\(30\)

Of course, if I didn't, I would not say that ;).

web2.0calc.com

whats the question

what have you tried urself

I am sure you are capable of doing something towards the answer.

Let the sides be x, y, z   and see what you come up with.

I think that that 7 in the 7y term looks incongruous.

I suspect a typo !

\(\begin{cases}725x + 727y = 1500 \text{ --- }(1)\\729x + 7y = 1508\text{ --- }(2)\end{cases}\)

Subtract (1) from (2) to get 

\(4x - 720y = 8\\ x - 180y = 2\text{ --- }(3)\)

Now, solve the system formed by equations (2) and (3) using method of elimination, and you will get \((x, y) = \left(\dfrac{271454}{131227}, \dfrac{50}{131227}\right)\)

Therefore, x - y = \(\dfrac{271404}{131227}\), as robloxDying suggested.

Hint: This looks complicated, but you can rewrite it into the system of inequalities \(\begin{cases}\dfrac t3 - 5 < t - 2 + 2t\\t - 2 + 2t \leq -3t + 7\end{cases}\), solve each inequality, and find the range of values of t which satisfies both inequalities. Now it just becomes two separate linear inequalities in t.

For l(x) to make sense, it is required that \(\dfrac{x - 1}{x - 3} \in (-1, 1)\).

Then, we solve the inequality \(-1 < \dfrac{x - 1}{x - 3} < 1\) to find the domain of f.

I will do one side of the inequality as an example. Let's say we want to solve \(\dfrac{x - 1}{x - 3} < 1\). We can't just multiply (x - 3) on both sides because it is not guaranteed that x - 3 is positive. Instead, we multiply (x - 3)^2 on both sides since it is always nonnegative. Also, note that x cannot be 3 because of the denominator.

\(\dfrac{x - 1}{x - 3} \cdot (x - 3)^2 \leq (x - 3)^2, \,x \neq 3\\ (x - 1)(x - 3) \leq (x - 3)^2, \, x \neq 3\\ x^2 - 4x + 3 \leq x^2 - 6x + 9,\, x \neq 3\\ 2x - 6 \leq 0, \, x \neq 3\\ 2x - 6 < 0\\ x < 3 \)

Now, you can solve the other inequality \(-1 < \dfrac{x - 1}{x -3}\) to get another inequality for x, combine them, and you get the domain of \(l(x)\).

 Thank you!!

13 x^2  + nx  - 17

We can factor  this  as

(13x  + 17) ( x - 1)   so   n  =  17 - 13  =  4

or

(13x - 1) ( x + 17)  so   n =  17*13 - 1 =  220

or

(x - 17) (13x + 1)    so n =  -17(13) + 1 =  -220

or

(x + 1) (13x - 17)    so   n =  -17 + 13  =   -4

The height of the trapezoid  can be found as

400 = (1/2) (20 + 30)(height)

height = 400 / 25  =   16

Triangles  AXB  and DXC    are similar

And   AB  : DC  =  2 : 3

Therefore   the    height of   triangle DXC  =  (3/5) 16  =  9.6

And the height of triangle  AXB  =  (2/5)16  = 6.4

So  the combined  areas of these triangles  = 

(1/2) ( DC * 9.6  + AB * 6.4)  = 

(1/2) ( 30 * 9.6  + 20 * 6.4)   =  208

And because of symmetry  triangles AXD and BXC  have the same areas

So,  triangle  BXC has an  area of  [ 400 - 208 ]  / 2 =  96

Happy birthday melody!!!

See the image below   (not to  scale)

The area of (1/2)  of the octagon =   (1/2) [ 2 * 12^2 ( 1 + sqrt 2)]  = 144 (1 + sqrt 2)

Triangle CDE  has an  area of   (1/2) CD * DE * sin 135°  =  (1/2) 12^2 sin 135° =   72 / sqrt 2 =  36sqrt (2)

So  the area of the trapezoid BCEF   =   

area of (1/2) octagon  -  area of triangle CDE

 144 ( 1 + sqrt 2)  - 36 sqrt 2  = 

[144 + 108sqrt 2] cm^2   ≈ 296.74 cm^2 

Thanks for that Tiggsy its great...its the solution I was looking for!! . Unknowingly I went for your second alternative ...only cos I had found a formula.

See the image below

Because  AB and DC are parallels cut by transversal AC, then  angles DAQ and NCQ  are equal alternate interior angles

And angle  PQD = angle CQN (vertical angles)

So...by AA congruency , triangle   AQD is similar to triangle CQN

But  NC =  (1/2) of  BC   so it also  equals (1/2)  of AD

Then NC :  AD  =  1 : 2

But this also means that  QN : QD = 1 : 2

So        QD  =     [ 2 / ( 1 + 2) ]  DN =    (2/3) (35) =    70 / 3