You are tasked with building a tower of blocks exactly 10 cm high. you have two types of block to work with. each block a is 2 cm high, and each block b is 3 cm high. you may use any number of each type of block, and may stack any kind of block on top of any other kind. how many possible towers are there

2 possible towers

Step-by-step explanation:

2 possible towers

Step-by-step explanation:

The height of the tower required is 10 cm, and that height may be obtained by 'x' number of blocks A and 'y' number of blocks, knowing its respective height, so 'x' and 'y' must accomplish the equation:

[tex] 10 = 2x + 3y [/ tex]

The maximum value for 'x' is 5; a higher value will give a higher than 10 cm height as a result. So, the possible answers for 'x' are: (0,1,2,3,4,5). Note that both 'x' and 'y' values ​​must be integers, and the only numbers that satisfy that condition are (x = 2, y = 2) and (x = 5, y = 0). The other options for 'x' will give a noninteger value for 'y' because of the equation above.

So, there are just two options.

See diagram, attached.

Part A: Equation of the directrix is  y= -2

Part B: point of intersection A of directrix & line of symmetry A(6,-2)

Part C: y =  (1/12) (x-6)^2+1

           or on simplifying

            12y = x^2 -12x +48

           (Note: given instructions omitted the +k term.)

part D: The parabola opens upwards becaus p>0.

Part E: The value of p is the difference of y coordinates of F minus the y-coordinate of A, all divided by 2, which is positive.

Part F: p = (4 - (-2))/2 = +3

Part G: y = (1/12)(x-6)^2 + 1

Part H: See attached diagram below

Part I : y = (1/12) (x-6)^2 + 1     where h=6, k=1, or V(6,1).

           It fits with the given formula, provided we include the +k term.

Step-by-step explanation:

The vertex form of the equation of a vertical parabola is given by y= 1/4p(x-h)^2 , where (h, k) is the vertex of the parabola and the absolute value of p is the distance from the vertex to the focus, which is also the distance from the vertex to the directrix. You will use the GeoGebra geometry tool to create a vertical parabola and write the vertex form of its equation. Open GeoGebra, and complete each step below. If you need help, follow these instructions for using GeoGebra.

Part A: Mark the focus of the parabola you are going to create at F(6, 4). Draw a horizontal line that is 6 units below the focus. This line will be the directrix of your parabola. What is the equation of the line? (y=-2)

Part B: Construct the line that is perpendicular to the directrix and passes through the focus. This line will be the axis of symmetry of the parabola. What are the coordinates of the point of intersection, A, of the axis of symmetry and the directrix of the parabola?  A(6,-2)

Part C: Explain how you can locate the vertex, V, of the parabola with the given focus and directrix. Write the coordinates of the vertex.

We know that the equation in vertex form is

y = (1/(4p)(x-h)^2+k   where V(h,k) are the coordinates of the vertex.

From points F and A, we know that the distance from directrix to focus is 6.

Therefore p=3.

The vertex, V(h,k) is located at the mid point between the segment joining the directrix and the focus, namely A & K, which has coordinates V(6,1), thus h=6, k=1.

The equation of the parabola is therefore

y = (1/(4p) * (x-h)^2 +k) = (1/(4*3))(x-6)^2 +1 = (1/12) (x-6)^2+1

(note: given instructions omitted the +k term.)

Simplifying

12y = x^2 -12x +36+12

12y = x^2 -12x +48

Part D: Which way will the parabola open? Explain.

The parabola opens upwards becaus p>0.

Part E: How can you find the value of p? Is the value of p for your parabola positive or negative? Explain.

The value of p is the difference of y coordinates of F minus the y-coordinate of A, all divided by 2, which is positive.

Part F: What is the value of p for your parabola?

p = (4 - (-2))/2 = +3

Part G: Based on your responses to parts C and E above, write the equation of the parabola in vertex form. Show your work.

y = (1/12)(x-6)^2 + 1

Part H: Construct the parabola using the parabola tool in GeoGebra. Take a screenshot of your work, save it, and insert the image below.

See attached diagram below

Part i: Once you have constructed the parabola, use GeoGebra to display its equation. In the space below, rearrange the equation of the parabola shown in GeoGebra, and check whether it matches the equation in the vertex form that you wrote in part G. Show your work.

1.33...x^2 - 16x -16y = -64

Multiply all by 3/4

x^2-12x-12y=-48

rearrange

12y = x^2-12x+36 +12

Factor

12y = (x-6)^2 + 12

Divide by 12 all through

y = (1/12) (x-6)^2 + 1     where h=6, k=1, or V(6,1).

It fits with the given formula, provided we include the +k term.