hectictar

The  x  coordinate of the vertex is the  x  value that makes whatever is in the absolute value be  0  .

So to find the  x  coordinate of the vertex, set what is inside the absolute value equal to  0 .

bx - 1  =  0      Now let's solve this equation for  x .

bx  =  1

x  =  1/b

To find the  y  coordinate of the vertex, plug in the  x  coordinate of the vertex into the function.

f( 1/b )  =  a | b(1/b) - 1 | + c

f( 1/b )  =  a | 0 | + c

f( 1/b )  =  c

So the vertex is  (1/b, c)

Take a look at this graph and try moving the sliders to help see it:

To get the inverse sin key, click the  " 2nd "  key in the top left corner, then click the  " asin "  key, which is right below it. Then enter your number and close parenthesee.

Or, you can type this into the box:    asin(2.23/3.78)    and hit enter.

You should get....

asin(2.23/3.78)  ≈  36.153°          or          asin(2.23/3.78)  ≈  0.631 radians

The side length of the cube  =  s

The number of square units in the surface area of the cube  =  6s²

The number of cubic units in the volume of the cube  =  s³

The number of square units in the surface area of the cube equals 1/6 of the number of cubic units in the volume. So...

6s² = (1/6)s³

36s² = s³

36  =  s³ / s²

36  =  s

The number of square units in the area of the square is equal to the number of cubic units in the volume of the cube. So...

The number of square units in the area of the square  =  s³

The side length of the square  =  √(the number of square units in the area of the square)

The side length of the square  =  √( s³ )

The side length of the square  =  s √s

The side length of the square  =  36 √36

The side length of the square  =  36 * 6

The side length of the square  =  216

Let  a  be the length of an edge of the tetrahedron.

A face of the cube looks like this:

By the Pythagorean theorem,

6² + 6²  =  a²

                            Combine like terms....  n + n = n * 2    so    6² + 6²  =  6² * 2

6² * 2  =  a²

                            Take the positive square root of both sides of the equation.

√[ 6² * 2 ]  =  a

√6²  *  √2  =  a

6√2  =  a

Since each face of the cube is the same, each edge of the tetrahedron is the same.

So the tetrahedron is a regular tetrahedron.

The formula for the volume of a regular tetrahedron is

volume  =  (edge length)³ / ( 6√2 )

                                                                   Plug in  6√2  for the edge length.

volume  =  ( 6√2 )³ / ( 6√2 )

                                                                   Simplify.

volume  =  ( 6√2 )( 6√2 )( 6√2 ) / ( 6√2 )

volume  =  ( 6√2 )( 6√2 )

volume  =  36 * 2

volume  =  72      cubic units

The radius of the first cone is 5 meters,

so the diameter of the first cone is 10 meters.

The diameter of the second cone is 15 meters.

$\frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{\text{diameter of second cone}}{\text{diameter of first cone}}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{15}{10}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\frac{27}{8}\\~\\ {\small\text{volume of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{volume of first cone}}$

Assuming that both haystacks have the same density, and each piece of hay weighs the same....

${\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{weight of first cone}}\\~\\ {\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot12 \text{ tons}\\~\\ {\small\text{weight of second cone}}\,=\,40.5 \text{ tons}$

I don't know the correct way to solve this, but first I guessed that x is 8 point something,

which makes  floor(x) = 8

And   70 / 8  =  8.75    , so this works:       $​​​​​​\lfloor 8.75 \rfloor \cdot 8.75 = 70$           

area of lower base  =  225π cm²

                                                                               and...

area of lower base  =  π(radius of lower base)²

                                                                               So...

π(radius of lower base)²  =  225π cm²

                                                                    Divide both sides of the equation by  π .

(radius of lower base)²  =  225 cm²

                                                                    Take the positive square root of both sides.

radius of lower base  =  15 cm

area of upper base  =  25π cm²

                                                                               and...

area of upper base  =  π(radius of upper base)²

                                                                               So...

π(radius of upper base)²  =  25π cm²

                                                                    Divide both sides by  π .

(radius of upper base)²  =  25 cm²

                                                                    Take the positive square root of both sides.

radius of upper base  =  5 cm

So this is what a cross section of the cone perpendicular to the base looks like:

where  h  is the altitude of the small cone that was cut off, and each length is in cm.

The bases of the frustrum are parallel, and corresponding angles are congruent,

So by AA similarity, we can say that the large right triangle and the small right triangle are similar.

So...

$\frac{\text{height of large triangle}}{\text{height of small triangle}}=\frac{\text{base of large triangle}}{\text{base of small triangle}}\\~\\ \frac{24+h}{h}=\frac{15}{5}\\~\\ \frac{24+h}{h}=3\\~\\ 24+h=3h\\~\\ 24=2h\\~\\ 12=h$

A point of intersection is a point that makes both equations true.

y  =  3x² + 4x - 5       and       y  =  x² + 11

Starting with the first equation...
 

y  =  3x² + 4x - 5

                                       Since  y  =  x² + 11  we can substitute  x² + 11  in for  y .

x² + 11  =  3x² + 4x - 5

                                       Subtract  x²  from both sides of the equation.

11  =  2x² + 4x - 5

                                       Subtract  11  from both sides of the equation.

0  =  2x² + 4x - 16

                                       Divide through by  2 .

0  =  x² + 2x - 8

                                       Factor the right side.

0  =  (x - 2)(x + 4)

                                       Set each factor equal to zero and solve for  x .

x - 2  =  0          or          x + 4  =  0

 x  =  2              or             x  =  -4

Use these values of  x  to find  y .

If    x  =  2    then    y  =  (2)² + 11  =  4 + 11  =  15    So   (2, 15)  is a point of intersection.

If    x  =  -4    then    y  =  (-4)² + 11  =  16 + 11  =  27    So   (-4, 27)  is a point of intersection.

The points of intersection are   (-4, 27)   and   (2, 15) .

y  =  ax² + bx + c

Since the graph passes through  (1, 10)  we know that when  x = 1 ,  y = 10 .

So we can replace  x  with  1  and  y  with  10  .

y  =  ax² + bx + c

10  =  a(1)² + b(1) + c

10  =  a + b + c

a + b + c  =  10  

The path from C to B must have two steps up and four steps right.

The possibilities are:

up, up, right, right, right, right

up, right, up, right, right, right

up, right, right, up, right, right

up, right, right, right, up, right

up, right, right, right, right, up

right, up, up, right, right, right

right, up, right, up, right, right

right, up, right, right, up, right

right, up, right, right, right, up

right, right, up, up, right, right

right, right, up, right, up, right

right, right, up, right, right, up

right, right, right, up, up, right

right, right, right, up, right, up

right, right, right, right, up, up

That is a total of  15  possibilities.