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alina1380 [7]
3 years ago
7

The parallelogram shown below has an area of 404040 units^2 2 squared.

Mathematics
1 answer:
TEA [102]3 years ago
8 0

Question:

The parallelogram shown below has an area of 40 units squared.

Find the missing base.

Answer:

See Explanation Below

Step-by-step explanation:

The question is incomplete as the diagram which shows the dimension of the parallelogram is not attached.

However, I'll give a general explanation on how to get the base of the parallelogram. If you follow this explanation, you'll get the right answer regarding your question.

The area of a parallelogram is calculated as thus.

Area = Base * Height

Given that the area = 40 units²

Let's assume the height of the parallelogram is 10 unit.

All you need to do is to plug in these values in the formula above..

Area = Base * Height becomes

40 = Base * 10

Divide both sides by 10

40/10 = Base * 10/10

4 = Base

Hence, Base = 4 units

Or take for instance the height is 5 units.

You'll still follow the simple steps as it is above.

Area = Base * Height becomes

40 = Base * 5

Divide both sides by 5

40/5 = Base * 5/5

8 = Base

Hence, Base = 8 units

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The correct statement comparing the theoretical and experimental probabilities is given as follows:

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Answer:

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\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

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