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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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riadik2000 [5.3K]
X=y+3
<span>-5x-4y=<span>30
-------------------
Substitute y + 3 for x in </span></span>-5x-4y=30
<span><span>-5<span>(y+3)-</span></span>4y</span>=<span>30
</span>-9y-15=<span>30
</span>----------------------------
Add 15 to each side
<span><span>-9y-15</span>+15</span>=<span>30+<span>15
-9y = 45
</span></span>-----------------------------------
Divide each side by -9
-9y ÷ -9 = 45 ÷ -9
y = -5
Now we solve for x
====================================================================
Substitute -5 for y in x = y + 3
x=-<span><span>5</span>+<span>3
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(-2,-5) is your answer


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3 years ago
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5

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2 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
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Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

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AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

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This means that you cannot find solutions of this equation. Then CD≠2 cm.

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