CPhill

√48  =

√ [16 * 3 ]  =

√ 16 * √3  =

4 √3

5.65  =

5 + 65 / 100  =

5 + (13 * 5) / (20 * 5)  =

5 +  13  / 20  =

[ 100 + 13]  / 20  =

113 / 20

15 x 15 x 15 x 15 x15  =  15⁵   = 759,375

10, 50, 5, 25 and 500  to make 575

One way :

500 - 25  + (10/5) * 50  =

475  + 2 * (50)  =

475 + 100 =

575

2^(x-1)+2^(2+x) =144    we can write       

2^x / 2  + 2^2 *2^x  = 144

2^x /2 + 4* 2^x = 144     multiply through by 2

2^x + 8*2^x  = 288

9*2^x  = 288    divide both sides by 9

2^x  = 32       write 32  as 2^5

2^x  = 2^5

So....x  = 5

(1/2)^4  =  (1) / 2^4  =    1/16

Good job, Gh0sty.....!!!!

Here's 22

f(x)  = -0.1x^2 + 3.2x -3.5

Let's  find the roots....thus

-0.1x^2 + 3.2x -3.5  = 0   multiply both sides by  -10

x^2 - 32x  + 35  = 0    subtract  35 from both sides

x^2  -  32x   =  -35       take 1/2 of 32  = 16....square this  = 256  ....add to both sides

x^2 - 32x + 256  = -35  +  256     factor and simplify

(x - 16)^2  = 221    take positive/negative roots of both sides

x - 16 =  ±√221      add 16 to both sides

x =   ±√221  + 16

So....the  two  roots are   -√  221  + 16  ≈  1.13   and √  221  + 16 ≈  30.87

And the positive difference between these two is the width of the tunnel at the bottom = 

30.87 - 1.13  ≈  29.74 ft....so....the tunnel is wide enough at the bottom

The function is an upside-down parabola....the vertex occurs at the max height

To find the height...let's find the x coordinate of the vertex.....this is given by

-b  / (2a)     where  b = 3.2   and a = -.1  ....so we have

-3.2 / [ 2(-.1) ] = -3.2 / -.2  = 32 / 2  =   16

So....putting this into the original function will give us the height....and we have

-.1(16)^2  + 3.2(16)  - 3.5  = 

-25.6 + 51.2 - 3.5   = 22.1  ft

However.......we need to find the width at 20 feet

So....set  the function = 20 and solve

20 =  -0.1x^2 + 3.2x -3.5  subtract 20 from both sides

Using the quadratic formula, we have that

x = ( -3.2 ±√ [ 3.2^2 - 4(-.1)(-23.5) ] ) / [ 2(-.1)] =

x = (-3.2 ±√ [  .84] ) / [ -.2 ]  =

x ≈ 11.41     and x  ≈ 20.58

So...the width at 20 feet  equals the distance between these ≈  9.17 feet

So....the tunnel is not wide enough at this point.....!!!!

2b^2 + b  = 6      

step 1....  divide both sides by 2

b^2  + (1/2)b   = 3        

step 2    .....take (1/2)  of (1/2)  = 1/4....square this = 1/16  add to both sides

b^2 + 1/2b + 1/16  = 3 + 1/16

step 3  ......factor the left side.....simplify the right side

(b + 1/4)^2  = 49/16

step 4  ......take positive/negative roots of both sides

b + 1/4  = ±√[ 49/16]

b + 1/4  =   ±7/4

step 5......subtract  1/4 from both sides

b = 7/4 - 1/4  = 6/4  = 3/2        or    b =  -7/4 - 1/4  = -8/4   = -2

The mistake was in step 2.....1/4 was added to both sides, but it should have been (1/4)^2 = 1/16 !!

Substituting x for t, this function is not one-to-one....thus....it has no inverse unless we restrict its domain

y = 100 ( 1 - t/40 )^2   divide both sides by 100

y / 100  = ( 1 - t/40)^2       take both roots of  sides

 ±√ (y/100)  = 1 - t/40       multiply both sides by -1

 ±√ ( y / 100 )  = t / 40 - 1     add 1 to both sides

1  ±  √ ( y / 100 )  = t/40        multiply both sides by 40

40  ± 40  √ ( y / 100 )  = t       swap t and y

40  ± 40  √ ( t / 100 )  = y    for  y, write  f^-1(t)

40  ± 40 √ ( t / 100 )  =   f^-1(t)

Here is the graph : https://www.desmos.com/calculator/lf5zoe7lei

If we restrict the original domain to ( -infinity, 40),  the inverse is represented by the orange graph

If we restrict the original domain to (40, infinity), the inverse is represented by the red graph