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The difference of 3 times the larger number x and twice a smaller number y is 95. The sum of 3% of the smaller and 5% of the larger is 5.7. Determine both numbers.

Hello Guest!

\(3x-2y=95\ |\ \times 3\\ \frac{3y}{100}+\frac{5x}{100}=5.7\\ \underline{5x+3y=570}\ | \times 2\)

\(\ 9x-6y=\ 285\\ \underline{10x+6y=1140}\)

\(19x=1425\)

\(x=75\)

\(3x-2y=95\\225-2y=95\\ 2y=130\)

y=65

  !

Just calc the individual areas and add them

rectangle area = l x w

     15x4    and   6x4

triangle area = 1/2 b h

    1/2  6 * 4     and     1/2   8 * 9

The graph of y=f(x) is shown in red below. Find f(2)

Hello Guest!

\(f(x)=-\frac{1}{2}x+3 \ |\ 0\le x\le 4\)

\(f(2)=2\)

  !

Sorry....   mislabeled those two angles in the diagram....

     switch them....

60 + 61 = 121

Use Pythagorean theorem to calculate the other side length:

61^2 = 60^2 + s^2
    Solve for 's'       Then add to 121 for perimeter

2) Axis of symmetry is x = 4.

sketch:

1) this is parabola in vertex form     vertex = (-3,-5)      axis of sym   x = -3

I answered a similar question, not the one asked. (Although the question has been changed to another similar question)

I answered like this because I did not believe the answer to the exact question would be the best way to help this student to learn.

He/she obviously lacks a primary school, pre-algebra, concept.  So I did not want to use algebra to answer.

Thanks Chris.  

You are welcome.

I should have seen the answer straight away but I didn't.   

Maybe I will next time....    

anyway, I am glad I could help.

1.5 x 12      x     2 x 12     x  452 =                 bacteria

thanks CPhill!

Use the angle of dec to the base of B to find the distance to B

tan 21.8 = 40/x        x = 100.00713 meters           (note the 21 degrees 48 min   is    21.8 decimal degrees)

Now use the angle of dec to the top

tan 12.5 = 40/x          x = 180.4283  meters     this is the distance line to the  to the triangle vertex which the top intercepts

    but the top is only    100.00713  meters away

(180.428-100.00713)/ 180.428   * 40 = 17.829  meters high = temple B

Check  tan 12.5 = 17.829/80.428 ....check !

\(\frac{y}{x\sqrt{2}}\)I might be wrong

Edit: I'm wrong sorry. IDK how to delete the answer.

\(x=1\quad \mathrm{or}\quad \:x=2\)

Thank you

LCM of 20 and 45  is 180min = 

   3 hours

6 Pm + 3 hours = 9 PM

Bus Bs route times will be:

45 mins

90 mins

135 mins

180 mins

Bus As times will be:

20 mins

40 mins

60 mins

...

180 mins.

Therefore 180 mins is the common factor, which is 9pm

Let  the cost of a soccer ball =  S

Let  the cost of a soccer bag = B

So

10S + 2B  =  155

12S  + 2B =  184        subtract  the first equation from the  second

2S =  29     divide both sides by  2

S  =29/2  =  $14.50    = the cost of a  soccer ball

So

10 (14.50)  + 2B   =155

145   + 2B   =155

2B  = 155   -145

2B  = 10

B  = $ 5  the cost of a soccer  bag

-sin x  = cos x      rearrange  as

cosx +  sin x   =    0

Note   :

at   3pi/4     sin =  sqrt (2)  /2      and   cos  =  -sqrt (2) /2

at  7pi/4      sin  = -sqrt (2)  /2    and  cos  = sqrt (2)  / 2

So   ...  3pi/4  and  7pi /4

sec^4 x  -  sec^2x   =

                                                      factor out sec^2 x

sec^2 x ( sec ^2 x   - 1)  =

                                                       { tan^2 x + 1   =  sec^2 x }

(tan^2 x + 1)  (tan^2 x)  =

tan^4 x  +  tan^2 x

Rearrange as

x^2 + (a + 8)x  +  20   =   0

Note  that 20  =

4 * 5

2 *10

1 * 20

So  

(x + 4) ( x + 5) =  x^2  + 9x  + 20    and  a  =1

(x +2)( x +10)  =x^2  +  12x + 20    and  a   = 4

(x + 20) ( x + 1)  = x^2  +  21x +  20    and  a =  13

This is just  a circle centered at the origin with a radius of  2

See here :  https://www.desmos.com/calculator/pdughwwuoo

(tan^2 x - 1) / sec^2 x     =      (tan x -cot x)  / ( tan x + cot x)

Working with the right side                          Working with the left side

sin x  /cosx    -  cos x / sin x                          tan^2 x   - 1   / ( 1 /cos^2x)

_______________________ 

sin x / cos x   +  cos x /sin x                          cos^2 x  ( sin^2 x / cos ^2 x   - 1)                   

(sin^2 x  - cos ^2x)  / (sin x cos x)

___________________________                   sin^2 x    -   cos ^2 x

(sin^2 x + cos^2 x) / (sin x cos x)

(sin ^2 x   - cos ^2 x)

_________________

               1

sin^2 x  -  cos ^2 x