Based on the calculations, we have the following:
- The area of the sheet of paper is 96 square inches.
- The combined area of the triangle cutouts is equal to 36 square inches.
- The area of the parallelogram is equal to 60 square inches.
- The altitude of the parallelogram is equal to 6.51 square inches.
<u>Given the following data:</u>
- Dimension of paper = 12-inch by 8-inch.
<h3>How to calculate the paper's area.</h3>
Mathematically, the area of the paper is given by this formula:

Area = 96 square inches.
<u>For the four (4) right triangles:</u>
- Dimension 1 = 2 inches by 9 inches.
- Dimension 2 = 3 inches by 6 inches.
Therefore, the combined area of the triangle cutouts is given by:

<h3>The area of the parallelogram.</h3>
This would be determined by subtracting the area of the four (4) right triangles from the areas of the paper as follows:

P = 60 square inches.
<h3>The altitude of the
parallelogram.</h3>

Altitude = 6.51 square inches.
Read more on parallelogram here: brainly.com/question/4459854
<u>Complete Question:</u>
A parallelogram is cut out of a 12-inch by 8-inch sheet of paper. There are four right triangle remnants. Two have the dimensions 2 inches by 9 inches, and the other two have the dimensions 3 inches by 6 inches. The resulting parallelogram has a base of approximately 9.22 inches.
The volume of HCl gas required to prepare the solution is 101.54 L
<h3>How to determine the mole of HCl </h3>
- Volume of solution = 1.25 L
- Molarity = 3.20 M
- Mole of HCl =?
Mole = Molarity x Volume
Mole of HCl = 3.2 × 1.25
Mole of HCl = 4 moles
<h3>How to determine the volume of HCl </h3>
- Number of mole (n) = 4 moles
- Temperature (T) = 30 °C = 30 + 273 = 303 K
- Pressure (P) = 745 torr = 745 / 760 = 0.98 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
Using the ideal gas equation, the volume of the HCl gas can be obtained as follow:
PV = nRT
Divide both side by P
V = nRT / P
V = (4 × 0.0821 × 303) / 0.98
V = 101.54 L
Learn more about ideal gas equation:
brainly.com/question/4147359
So using a(2)=0 we can first solve for k by substituting t for 2
0 = (2-k)(2-3)(2-6)(2+3)
0 = (2-k)(-1)(-4)(5)
0 = (2-k)20
0 = 40 - 20k
-40 = -20k
k = 2
The next step would be to find all the 0s of a.
0 = (t-2)(t-3)(t-6)(t+3)
T = 2,3,6,-3
Then we find the product
2x3x6x-3 = -108
Since the problem asks for the absolute value, the answer is positive 108
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