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Sedaia [141]
3 years ago
12

A rectangular pool is surrounded by a walk 4 meters wide. The pool is 5 meters longer than its width. If the total area of the p

ool and walk is 488 square meters more than the area of the​ pool, find the dimensions of the pool. The width of the pool is ____m and the length of the pool is____ m.
Mathematics
2 answers:
larisa [96]3 years ago
6 0

Answer:

45 and 55

Step-by-step explanation:

emmasim [6.3K]3 years ago
4 0

Answer: The width of the pool is 24m and the length of the pool is 29m

Step-by-step explanation: The dimensions of the rectangular pool is not given, but the clues we have are that the length of the pool is 5 meters longer than it’s width. With that information, we can say that the width is W, while the length is W + 5.

Also the pool is surrounded by a walk 4m wide. That means the total length including the two walkways is W + 5 + 4 + 4 (that equals W + 13), and the total width including the two walkways is W + 4 + 4 (that equals W + 8). We have yet another clue, which states that the total area of the pool and walk is 488m more than the area of the pool.

We shall label the area of the pool as AP. If Area = L x W, then area of pool is

AP = (W + 5) x W

AP = W^2 + 5W

Also the total area of the pool and walk shall be labeled as APW. Therefore,

APW = (W + 13) x (W + 8)

APW = W^2 + 8W + 13W + 104

APW = W^2 + 21W + 104

Having calculated the area of the pool as W^2 + 5W and the total area of the pool and walk as W^2 + 21W + 104, and remembering that the area of pool and walk is 488 meters more than the area of the pool, we can now write the following equation;

W^2 + 21W + 104 = W^2 + 5W + 488

Collecting like terms we now have

W^2 - W^2 + 21W - 5W = 488 - 104

16W = 384

Divide both sides of the equation by 16

W = 24.

If the length of the pool has been given as 5 meters more than its width, then length of the pool equals 24 + 5 and that equals 29 meters. Therefore

Length = 29m and Width = 24m

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Find the 17th term in the sequence for which a1 = 10 and d = -3.
yKpoI14uk [10]

Answer:

d. -38

Step-by-step explanation:

Arithmetic sequence concepts:

The general rule of an arithmetic sequence is as follows:

a_{n+1} = a_{n} + d

In which d is the common diference between each term.

We can expand the general equation to find the nth term from the first, by the following equation:

a_{n} = a_{1} + (n-1)*d

In this question:

a_{1} = 10, d = -3

17th term:

a_{17} = a_{1} + (17-1)*d = 10 + 16*(-3) = -38

So the correct answer is:

d. -38

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3 years ago
a bag contains 5 blue marbles,3 red marbles, and 2 yellow marbles. You select a marble at random. What is p (yellow)
Lunna [17]

Answer:

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Answer:

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Step-by-step explanation:

42=8m+13m

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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] =<em>I</em>.

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

<em>I</em> = \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
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