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3 years ago

The shape of the space is a rectangle. What is the actual area of the space they are filling with cement?

Mathematics

1 answer:

Leni [432]3 years ago

4 0

Answer:

The area would be the length of the rectangle times the with of the rectangle.

Step-by-step explanation:

Imagine the rectangle is cut up into tiny squares. For example, if the length of the rectangle is 5, and the with is 6, than each square would represent 1 unit that is in the area of the rectangle. So if you put 6 squares on the length of that rectangle, that side would be filled up, because the with is 6 and the length would be 4. Add 6 more squares on the with and you have 3 more square colluoms. Add 3 more times 6 than the rectangle would be filled with squares. What we just did is finding the area, which can be simplified to making 5 rows of 6 unit square colloums. This can be even more simplified by the equation length times with

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What is -2xm = 1, for x

Evgesh-ka [11]

Answer:

x = 1/2m

Step-by-step explanation:

4 0

3 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago

Find the Sum (3x+x^7 +5x^2) +( 4x^2 -2x +8).

dexar [7]

Answer:

x^7 + 9x^2 + x + 8

Step-by-step explanation:

Combine like terms

5 0

3 years ago

I’m stuck on this I would really appreciate if someone could answer this and explain so I could understand better :)

BARSIC [14]

Complementary angles have a sum of 90 degree, right angle
It would be C. Since it’s the only 2 angles that form a right angle

4 0

3 years ago

What is the area of this rectangle

MrRissso [65]

Answer:

6 cm^2

Step-by-step explanation:

Area of Rectangle is L*W

L=3 and W=2

3*2=6

4 0

2 years ago