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Rainbow [258]
2 years ago
7

If the length football field is 100 yards and the width is 53.3 yards, what is the area?

Mathematics
2 answers:
riadik2000 [5.3K]2 years ago
4 0
Formula for area of a rectangle is: A = l x w
A: Area
l: length
w: width

length of football field is 100 yards and width is 53.3 yards.
To find the area we need to multiply the length by width.

Area of football field = 100 x 53.3 = 5330 square yards (yards^2)




Xelga [282]2 years ago
3 0
The area is 5330 yards
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Answer:

Matrix multiplication is not conmutative

Step-by-step explanation:

The matrix multiplication can be performed if the number of columns of the first matrix is equal to the number of rows of the second matrix

Let A with dimension mxn and B with dimension nxp represent two matrix

The multiplication of A by B is a matrix C with dimension mxp, but the multiplication of B by A is can't be calculated because the number of columns of B is not the number of rows of A. Therefore, you can notice that is not conmutative in general.

But even if the multiplication of AB and BA is defined (For example if A and B are squared matrix of 2x2) the multiplication is not necessary conmutative.

The matrix multiplication result is a matrix which entries are given by dot product of the corresponding row of the first matrix and the corresponding column of the second matrix:

A=\left[\begin{array}{ccc}a11&a12\\a21&a22\end{array}\right]\\B= \left[\begin{array}{ccc}b11&b12\\b21&b22\end{array}\right]\\AB = \left[\begin{array}{ccc}a11b11+a12b21&a11b12+a12b22\\a21b11+a22b21&a21b12+a22b22\end{array}\right]\\\\BA=\left[\begin{array}{ccc}b11a11+b12a21&b11a12+b12a22\\b21a11+b22ba21&b21a12+b22a22\end{array}\right]

Notice that in general, the result is not the same. It could be the same for very specific values of the elements of each matrix.

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Step-by-step explanation:


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Answer:

a) Let \frac{a}{b}=\frac{-1}{x^2}, \text{ and } \frac{c}{d}=\frac{1}{x}.

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b)

Let \frac{a}{b}=\frac{1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x^2}.

Observe that

\frac{a}{b}-\frac{c}{d}=\frac{1}{x}-\frac{1}{x^2}=\frac{x^2-x}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

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Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x}.

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