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'A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. The equations in thesystem can be linear or non-linear.'                                                                     --http://www.sosmath.com/soe/SE/SE.html

1 Minute = 60 Seconds

5 Minutes = X Seconds

You can find X by multiplying 60 by 5.

A right triangle is a triangle with a 90 degree angle in it

cameron has budgeted $240 on meals for his full trip. 36.8% transportation,32% lodging,meals 19.2%.and gifts 12%.what is the total budget for the trip?

  total budget  =    240 / 19.2%  =  240 / .192  = $1250

Oops, I made a typo when I was submitting the problem (sorry!). 

"AX = 3 CX = 12" should actually be AX = 3CX = 12. 

4.) Find the side of a rhombus whose diagonals are 10cm and 26cm.

Side  = √[5^2 + 13^2]  = √ [25 + 169]  = √ 194  cm  ≈  13.93 cm

If you plug the problem you showed straight into a calculator with no parenthesis, you get 87.0351681957186544.

19. Also, some people don't like it when people ask this question, because it gets really repetitive and annoying. Don't ask this question again, please.

Note higgsb, we don't have enough info  to  answer this

-1/8+3/4

-1/8+6/8

6/8-1/8

=5/8

52.5

The answer is 2,3, and 4 . 

3.) A man travels 16km north and the 8km west. How far is he from the starting point?

√[ 16^2  + 8^2]  = √[256 + 64]  = √ 320  = 8√5 km ≈ 17.89 km

5,) The length of a rectangle is 7cm more than its width. If a diagonal measures 17cm. Find the dimensions of the

W^2  +  [ W + 7]^2  = 17^2

W^2  + W^2  + 14W  +  49  = 289

2W^2  + 14W  − 240  = 0

W^2 + 7W  −  120  = 0

(W − 8) (W + 15)  = 0

W − 8  = 0       →   W  = 8cm   and L = W + 7  = 8 + 7  = 15cm

7.) A right triangle has legs a and b and hypotenuse c. Prove that the altitude h to the hypotenuse is h = ab / c.

The area =  a * b / 2   =   h * c / 2   →    a*b  = c*h  →   h =  ab / c

1.) Triangle ABC has a right triangle at C. CD is the altitude to AB. If CD = 6cm, AD = 3cm and DB= 5x - 3cm. Find x, AB, AC and BC.

ΔADC  ~ ΔACB

AD/DC  = AC/CB

AC  =  √[AD^2 + DC^2 [  = √[ 3^2 + 6^2]  = √45 = 3√5   ...so....

AD/DC  = AC/CB

3/6  = [3√5] / [6√5]       so  BC  = 6√5  = √180

AC^2  + BC^2  = AB^2

45   + 180  =  AB^2

225  = AB^2

15  = AB

AD + DB  = 15

3 + DB  = 15  →   DB   = 12

DB = 12  =  5x - 3  →   15  = 5x  →  x  = 3

7-15 =  - 8    (really !)   

what is 7-15

Very nice, EP......!!!!!

Remove the ending 5 from 365  = 36

Double this and add1 =   36 * 2  + 1  =  72 + 1  = 73

And that's  how many times can 5 go into 365.....!!!!

Let 'x' be the number

15 + 8x = 7    SUbtract 15 from both sides of the equation

8x = -8           Divide by 8

x = -1

Let  one side of the rectangle = x     ....and the other side  =  1.25x

Then the  perimeter  =  2 [ 1x + 1.25x]  = 2 [2.25x] = 4.5x

So the side of the square  =  [4.5 x ] / 4  

And the area of the rectangle  =  x * 1.25x  = 1.25x^2    (1)

And the area of the square  = ( [4.5 x ] / 4)^2  = [ 20.25/16] x^2 = [81/64]x^2    (2)

So the ratio  of  (1) to (2)   =   1.25x^2 / [ (81/64)x^2 ] =  1.25 / (81/64)  = [5/4] / [ 81/64]  =

[5/4] * [ 64/81]  =     [64/4] * [ 5/81]  =  [16*5] / 81  =  80/81    as EP  found!!!

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