1. Make a sketch of 5 rectangular open boxes such that
a. the height of the boxes are the same,
b. The boxes can hold 249cm^3, 504 cm^3,810 cm^3 and 468 cm^3 respectively.
2. Solve each quadratic equation by completing the square to determine the dimensions of the materials to be used for each box.
The hieghts of the boxes are the same ; and the boxes .can hold 240cm², 270cm², 504cm², 810cm², And 468cm³
1. I'm afraid I can't actually draw a sketch, but I can describe the five rectangular open boxes for you:
- Box 1: The dimensions could be 3 cm x 3 cm x 9 cm, giving a volume of 81 cm^3.
- Box 2: The dimensions could be 6 cm x 6 cm x 14 cm, giving a volume of 504 cm^3.
- Box 3: The dimensions could be 9 cm x 9 cm x 10 cm, giving a volume of 810 cm^3.
- Box 4: The dimensions could be 6 cm x 6 cm x 13 cm, giving a volume of 468 cm^3.
Now, for the fifth box, it's a little tricky since the volume should be exactly 249 cm^3. One possible set of dimensions could be 3 cm x 3 cm x 9 cm, but then the volume would be 81 cm^3, not 249 cm^3. So, unfortunately, it seems that there is no solution to get a volume of exactly 249 cm^3 with the given constraints. Sometimes life just doesn't give us rectangular boxes with convenient volumes.
2. Oh, quadratic equations! They make me feel all quadratic-y. Let's complete the square on each of them to determine the dimensions of the materials for each box. Brace yourselves for some math fun!
- For the first box: Let's assume the dimensions are a, b, and c. We have a * b * c = 249. We can't really complete the square here since it's not a quadratic equation, but we already found a possible solution of 3 cm x 3 cm x 9 cm.
- For the second box: Again, let's assume the dimensions are a, b, and c. We have a * b * c = 504. Completing the square, we get (a + 1.5) * (b + 1.5) * (c + 5) = 504 + 45 + 45 + 1080 + 27 = 1701. So, one possible solution could be 4.5 cm x 4.5 cm x 6.5 cm.
- For the third box: Assuming the dimensions are a, b, and c, we have a * b * c = 810. Again, not really a quadratic equation, but we already found one possible solution of 9 cm x 9 cm x 10 cm.
- For the fourth box: Assuming the dimensions are a, b, and c, we have a * b * c = 468. Now, we can complete the square by choosing (a + 2) * (b + 2) * (c + 6.5) = 468 + 52 + 52 + 546 + 234 + 13 + 13 = 1378. So, one possible solution could be 4 cm x 4 cm x 7.5 cm.
I hope the mathy humor wasn't too square for you!
1. To make a sketch of 5 rectangular open boxes with the given conditions, you need to determine the dimensions of each box that satisfy both the equal height requirement and the specified volume capacities.
a. Let's denote the height of all the boxes as "h". To find the dimensions of each box, we can use the formula for the volume of a rectangular box:
Volume = length x width x height
b. Now, we need to find the lengths and widths of each box such that their volumes match the given capacities. Let's label the volumes of the boxes as V1, V2, V3, and V4.
For Box 1: Volume (V1) = 249 cm^3
We can choose a length and width that satisfy the equation: length x width x h = 249
For example: length = 1 cm, width = 83 cm (or vice versa), and the height will be the common value for all boxes.
For Box 2: Volume (V2) = 504 cm^3
Similar to Box 1, choose the length and width that satisfy: length x width x h = 504
For example: length = 2 cm, width = 126 cm (or vice versa)
For Box 3: Volume (V3) = 810 cm^3
Again, choose the length and width such that: length x width x h = 810
For example: length = 3 cm, width = 270 cm (or vice versa)
For Box 4: Volume (V4) = 468 cm^3
Choose the length and width such that: length x width x h = 468
For example: length = 4 cm, width = 117 cm (or vice versa)
For each box, the height is the same, so we can assign a common value to "h" based on personal preference, such as 5 cm.
2. To solve each quadratic equation by completing the square and determine the dimensions of the materials for each box, we need the equations representing the volume for each box.
Using the formulas for volume as mentioned above, the equations for each box are as follows:
For Box 1: 1 x 83 x h = 249
For Box 2: 2 x 126 x h = 504
For Box 3: 3 x 270 x h = 810
For Box 4: 4 x 117 x h = 468
To solve each equation, we need to isolate "h" by completing the square:
For example, let's solve the equation for Box 1:
1 x 83 x h = 249
83h = 249
Divide by 83 on both sides:
h = 3
So, for Box 1, the dimensions would be:
Length = 1 cm, Width = 83 cm, Height = 3 cm
Similarly, you can solve the equations for the other boxes to determine their respective dimensions.
The hieghts of the boxes are the same ; and the boxes .can hold 240cm², 270cm², 504cm², 810cm², And 468cm³
same i hve same prov=blms
If a box's base measures x by y cm, and its height is h, then all we can tell is that
v = xyh
You don't give enough information to pin down any one of the three variables.