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777dan777 [17]
3 years ago
15

2.Solve the following quadratic equations i. 9x^2 - 1/16 ii. 2h^2 - 3h - 27

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
4 0
i)~9x^2-\dfrac{1}{16}=0\iff9x^2=\dfrac{1}{16}\iff x^2=\dfrac{1}{9\cdot16}\iff \\\\x^2=\dfrac{1}{144}\iff x=\pm\sqrt{\dfrac{1}{144}}=\pm\dfrac{1}{12}\Longrightarrow\boxed{x=\pm\dfrac{1}{12}}


ii)~2h^2-3h-27=0\\\\\Delta=b^2-4ac\to \Delta=(-3)^2-4\cdot2\cdot(-27)\to\Delta=9+216=225\\\\
\Longrightarrow h=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-(-3)\pm\sqrt{225}}{2\cdot2}=\dfrac{3\pm15}{4}\\\\\begin{cases}h_1=\dfrac{3+15}{4}=\dfrac{18}{4}\iff \boxed{h_1=\dfrac{9}{2}}\\h_2=\dfrac{3-15}{4}=\dfrac{-12}{4}\iff\boxed{h_2=-3}\end{cases}

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Answer:

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Step-by-step explanation:

The first cave has 7 times more bats than the last cave.  So if the 45th cave has b bats, then the first cave has 7b bats.

There are 77 bats in every row of 7 caves.  So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.

Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats.  Repeating this logic:

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#23-28 = 77−7b

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So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.

Now we do the same thing from the other end.  If cave #45 has b bats, then caves #39-#44 have 77−b bats.  And since caves #38-44 have 77 bats, then cave #38 has b bats.  Therefore:

#45 = b

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#38 = b

#32-37 = 77−b

#31 = b

So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.

Adding that to the first 29 caves, plus x number of bats in cave #30:

308 + 7b + x + 154 + b = 462 + 8b + x

We know this equals 490.

490 = 462 + 8b + x

28 = 8b + x

x is a maximum when b is a minimum, which is b = 2.

28 = 8(2) + x

x = 12

There are at most 12 bats in the 30th cave.

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