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3 years ago

Explain how the equation A=lXw

Mathematics

2 answers:

Margaret [11]3 years ago

5 0

Its how u find the perimeter of rectangles and squares

Ratling [72]3 years ago

3 0

No, its how you find the area of mainly rectangles and squares. A=LxW can help lead to more formulas of triangles and other shapes.

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First twelve mulitples of 4

Pachacha [2.7K]

Answer:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

Step-by-step explanation:

4*1 = 4

4*2 = 8

4*3 = 12

and so on...

5 0

2 years ago

Read 2 more answers

A team pays $50 to reserve a banquet table. The players on the team pay

Margaret [11]

Answer:

6 players

Step-by-step explanation

Since the team has to pay $50 to reserve a table, you need to subtract this fixed cost from $104. Then you divide the total variable cost, $54, by what each player has to pay, $9, to find how many players sit at the team table.

6 0

2 years ago

The angle of ABC is how many degrees

Goshia [24]

Answer:

I think the awsner is 180 deegrees

7 0

2 years ago

Which reason could be used for statement 3??

Basile [38]

Option 3, definition of congruent segments is the answer I'm pretty sure

4 0

2 years ago

M1 Matrix is:

harkovskaia [24]

Answer:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

Step-by-step explanation:

Given matrix $M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

The Inverse of a 2X2 matrix

$A=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)$

can be found using the following:

$A^{-1}=\dfrac{1}{ad-bc} \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)$

Therefore:

$M^{-1}=\dfrac{1}{(5*-5)-(3*-8)} \left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=-1\left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

Next, we find the product $M^{-1}M$

$M^{-1}M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\\=\left(\begin{array}{ccc}-5*-5+3*-8&-5*3+3*5\\-8*-5+5*-8&-8*3+5*5\end{array}\right)\\=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

Therefore:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

NOTE: The multiplication of a matrix and its inverse always gives the identity matrix as seen above,

7 0

3 years ago