CPhill

Manuel.... I made a typo on this question earlier....see my corrected answer here...

https://web2.0calc.com/questions/help-please_26975#r1

[ I verified my result with WolramAlpha...]

Leonhard is doing a fundraiser for his school to buy AoPS books by running a lemonade stand. At the beginning of the day, Leonhard has 25 dollars of cash in his box. In the morning, he sells 8 dollars of lemonade, but he has to go to the grocery store and buy 5 dollars worth of sugar. At lunchtime, he sells 15 dollars worth of lemonade. In the evening, he has to go buy more lemons and the wind blows 3 dollars out of his box. The lemons cost 6 dollars. However, he also sells 25 dollars worth of lemonade during the evening. If Leonhard uses the money from his box to make his purchases, how much money should he have in his box at the end of the day?

He starts with $25.....to this he adds  $8 + $15 + $25   =  $73

From this....he subtracts  $5 + $3 + $6  = $14

So.....he should end up with  $73 - $14   =  $59

4.)

In triangle $ABC$, $m\angle A = 90^\circ$, $m\angle B = 75^\circ$, and $BC = \sqrt {3}$ units. What is the area of triangle $ABC$? Express your answer as a common fraction.

C

                 √3

A   90°           75°   B

sin 75°  =  AC / √3

AC  = √3* sin (75°)

Angle ACB  =  90 - 75  =  15°

sin(15°)  = AB / √3

AB  =  √3 * sin (15°)

So   the area  is

(1/2) AB * AC   =

(1/2) (√3 sin(15°) * √3 sin (75°)   =  (3/2)sin(15°)sin(75°)  =

(3/2)sin(15”) sin ( 90° - 15°)  =

(3/2)sin(15°)  (sin90”cos15°  - cos90°sin15°)  =

(3/2)sin(15°) (cos (15°) =        

(3/2)(1/2) sin (2 * 15)  =

(3/2)(1/2)sin(30°) = 

(3/2)(1/2)sin (30°)  =

(3/2) (1/2)(1/2)  =  3/8  units^2

4√15 - 10√3/5 + 3√60  =

4√15  - 2√3  +  3√[ 4 * 15]  =

4√15  - 2√3  + 3 [ √4  * √15 ]  =

4√15  - 2√3  + 3*2 *√15  =

4√15  - 2√3  + 6√15  =

10√15 - 2√3  

Edit to correct a typo

Angle AFB  =  Angle DFE

Angle DBA  = Angle FDE

Therefore, by AA congruency  Triangle DEF  is similar to Triangle BAF

But  DE  is  2/3  of DC.... and since AB  = DC, then  DE  is also 2/3 of AB

Then all  parts of triangle DEF  are 2/3  of  their corrresponding parts in triangle BAF

Thus....the altitude of triangle  DEF  is 2/3  of the altitude of triangle BAF

But the altitude of triangle DEF  = CG since they are parallel

And for the same reason, the altitude of triangle BAF   = BG

Thus  CG   =  2/3  of BG.....so ...  there are 5 parts of BC......and BG  is 3 of these

So  BG  =  3/5   of BC

And triangle BGF  is similar to  triangle BCD

But  BG  =  3/5  of BC....so  GF  =  3/5 of CD

So    GF  = FG  =   (3/5)  20    =   12

1 dgit numbers  = 4

2 digit numbers  = take any 2 of 4 numbers and permute them  =  P(4,2)  = 12

3 digit numbers  =   take any 3 of  4 numbers and permute them = P(4,3)   = 24

4 digit numbers  = permute all 4 numbers =  4!  =  24

So.....

4  + 12  +  2(24)    =     64 numbers 

Another approach here without resorting to Trig is to just use similar triangles.

We have that

BC / CA   =   EF / FA

x / 28  =  5 / 7        multiplt both sides by 28

x  =  (28/7) *  5   =   4 * 5   =    20  ft

Good job, ACG......!!!....these Geometry problems are tough, sometimes...!!!

Manuel....I guess we have this :

³√ 5   ³√ 4

But......I'm not sure what you want to do with these.....we can't add them because we don't have the same quantities under the radicand

Multiplying them we get

³√20

I'm confused....!!!

Look at the secant graph here :

https://www.desmos.com/calculator/g2ulbffk6j

Note  that  the range  is   ( - inf, -1 ] U [ 1, inf )

So.....the secant's  range  always has an absolute value   ≥  1

Only the first answer  meets that criteria....so....you are correct  !!