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beks73 [17]
3 years ago
6

In a small Library there are five non-fiction books for every nine fiction book there are 130 Non-fiction books how many fiction

books are there
Mathematics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

For 130 non-fiction books, the number of fiction books can be from 270 to 278.

Step-by-step explanation:

Given that, there are 5 non-fiction books for every 9 fiction books.

Observe that, if the number of fiction books is less than 9, than there will be no no-fiction books.

So, there are 5 non-fiction books for every 9 to (9+8) fiction books

So, by elementary method, there will be 5n non-fiction books for every 9n+8 fiction books, where n is a natural number.

Or, if the number of non-fiction books is 5n, then the number of fiction books will be ranging from 9n to (9n+8).

For, 150 non-fiction books, 5n=150 or n=150/5=30.

Hence, the number of fiction books can be from 9n to (9n+8).

=9x30 to (9x30+8)

=270 to 278.

Hence, for 130 non-fiction books, the number of fiction books can be from 270 to 278.

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Jerry was using matrices to solve the system of three equations. He has shown all his steps. Did he make a mistake if so in what
melomori [17]

There are several ways to solve a system of equation; one of these ways is the use of matrix.

<em>Jerry made a mistake at step 2</em>

From the attachment, the step 1 is represented as:

\mathbf{\frac 12R_1 \left[\begin{array}{cccc}1&1&1&0\\5&3&-2&-4\\3&2&1&1\end{array}\right] }

The equation in step 2 is given as:

\mathbf{R_2 = 5R_1 - R_2}

This means that:

We subtract the elements of row 2 from the elements of row 1, multiplied by 5

So, we have:

\mathbf{R_2 = 5\left[\begin{array}{cccc}1&1&1&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }

Expand

\mathbf{R_2 = \left[\begin{array}{cccc}5&5&5&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }

Subtract corresponding cells

\mathbf{R_2 = \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }

So, the new row 2 elements should be:

\mathbf{ \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }

However, the row 2 elements of Jerry's steps are:

\mathbf{R_2 = \left[\begin{array}{cccc}0&-2&-7&-4\end{array}\right] }

This means that:

Jerry made a mistake; and the mistake is at step 2

Read more about matrix at:

brainly.com/question/21848291

4 0
2 years ago
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The widths of two similar rectangles are 10 m and 15 m. What is the ratio of the perimeters? Of the areas?
vitfil [10]
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3 years ago
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JulsSmile [24]
<h3>Answer:  25w+200 > 750</h3>

==========================================================

Explanation:

He starts off with 200 cards. Then he adds on 25w more cards for each week (w). Overall, he'll have 200+25w cards

We can think of it like this:

  • After 1 week, he adds on 25*1 = 25 cards
  • After 2 weeks, he adds on 25*2 = 50 cards total
  • After 3 weeks, he adds on 25*3 = 75 cards total
  • After 4 weeks, he adds on 25*4 = 100 cards total, and so on.
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So that's another way to see where the 25w comes from.

The expression 200+25w is the same as 25w+200. This is because we can add two numbers in any order.

------------

Since he wants to know when he'll have more than 750 cards, this means we set 25w+200 greater than 750.

That's how we get to the answer of 25w+200 > 750

Notice how there isn't a line under the inequality sign. We aren't using the "greater than or equal to" symbol here. We want to know when the cards gets over 750, but we don't want to know when it's equal to 750.

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