hectictar

Given    \(f(x)=3x^2+10x-8\)    and    \(g(x)=3x^2-2x\)

What is \((\frac{f}{g})(x)\)   ?

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\((\frac{f}{g})(x)\,=\,\frac{f(x)}{g(x)}\)

                                              Substitute  \(3x^2+10x-8\)  in for  \(f(x)\)  and substitute  \(3x^2-2x\)  in for  \(g(x)\)

\((\frac{f}{g})(x)\,=\,\frac{3x^2+10x-8}{3x^2-2x}\)

                                              Split  10x  into two terms such that their coefficients add to  10  and multiply to  -24

\((\frac{f}{g})(x)\,=\,\frac{3x^2+12x-2x-8}{3x^2-2x}\)

                                              Factor the numerator and denominator.

\((\frac{f}{g})(x)\,=\,\frac{3x(x+4)-2(x+4)}{3x^2-2x}\\~\\ (\frac{f}{g})(x)\,=\,\frac{(x+4)(3x-2)}{x(3x-2)} \)

                                              Cancel the common factor of  (3x -2)  and note that  3x - 2 ≠ 0 , that is,  x  ≠  2/3

\( (\frac{f}{g})(x)\,=\,\frac{x+4}{x}\qquad\text{and}\qquad x\neq\frac23\)

Since  x  is in the denominator we can also note that  x ≠ 0

If  A = {2, {4, 5}, 4}, true or false:

Our answers are different on 1 and 3.

Also...I don't know if the ⊂ symbol means a proper subset or just a subset, but it doesn't really matter for these questions. 

2.

What is the parent function and what general shape does the parent function have?

The parent function is  f(x)  =  |x|

Its general shape is a "V" shape.

Explain all of the parameters and what they do to change the graph.

The graph of  f(x)  =  -3|x + 1| - 2  is the graph of  f(x)  =  |x|

shifted to the left by  1  unit,

stretched vertically by a factor of  3,

flipped over the x-axis,

and shifted down by  2  units.

1  is added to x, which shifts the graph to the left by  1  unit.

3  is multiplied by  |x + 1|  which stretches the graph vertically by a factor of  3 .

-1  is multiplied by  3|x + 1|  which flips the graph over the  x-axis.

-2  is added to -3|x + 1|  which shifts the graph down by  2  units.

1.

Graph the function  \(f(x)=\begin{cases} x+6 & x\leq -4 \\ (x+2)^2-3 & -4< x\leq 1\\ -|x-4|+5 & x>1 \end{cases} \)

Here's the graph of the function on desmos:

AHH Really? Yay!!!

I counted  42  different planes.

Also, since for example, the plane that goes through ABC also goes through the points  E  and  D, I called that plane ABCDE.

I named each plane by each vertex on the prism that it contains.

*edit* This could be wrong! I am not totally sure about it.

We can arrange them like this so there is only one unused square:

I could not figure out how fit them together so there were no unused squares.

After reading about "tetrominoes" on Wikipedia, I found out that the 5 pieces above are the 5 "free tetrominoes," and "Although a complete set of free tetrominoes has a total of 20 squares, they cannot be packed into a rectangle." See:

This is only for the second question:

We can multiply each dimension by  4  to make each dimension an integer. This will only scale the prism and the cubes up by  4....it won't change the number of cubes that can be fit into the prism.

So the new dimensions of the prism are  18 ft.  ×  30 ft.  ×  45 ft.

We can't cut the prism into cubes with sides length 18 ft, because 18 does not fit evenly into  30.

We can't cut the prism into cubes with sides length  9  ft, because  9  does not fit evenly into  30.

We can't cut the prism into cubes with sides length  2  ft, because  2  does not fit evenly into  45.

The side length of the cubes must fit evenly into 18, 30, and 45.

That is, the side length of the cubes must be a factor of 18, 30, and 45.

The maximum side length of the cubes is the greatest common factor of 18, 30, and 45.

The maximum side length of the cubes is  3 (ft.)

After splitting the prism into cubes with sides length 3 ft,

the prism will be 6 cubes high, 10 cubes wide, and 15 cubes deep.

the total number of cubes  =  6 * 10 * 15  =  900

a)

x  =  2√2 cos θ

y  =  2 sin θ

x  =  2√2 cos θ

                            Square both sides of this equation

x²  =  8 cos² θ

                            Divide both sides by  8.

x² / 8  =  cos² θ

y  =  2 sin θ

                                   Square both sides of this equation.

y²  =  4 sin² θ

                                   By the Pythagorean identity, we can substitute  (1 - cos²θ)  in for  sin²θ

y²  =  4(1 - cos²θ)

                                   Substitute  x² / 8  in for  cos²θ

y²  =  4(1 -  x² / 8)

                                   Divide both sides of the equation by  4 .

y² / 4  =  1 - x² / 8

                                   Add  x² / 8  to both sides of the equation.

x² / 8  +  y² / 4  =  1

Here is what the ellipse looks like: