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uranmaximum [27]
3 years ago
11

There is a mountain with 45 bat caves in a row. Every cave has at least 2 bats and there are 490 bats in all. Any 7 caves in a r

ow contain exactly 77 bats. Suppose the first cave has 7 times more bats than the last cave. What is the greatest possible number of bats in the 30th cave?
Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

12

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:

#1 = 7b

#2-#7 = 77−7b

#8 = 7b

#9-#14 = 77−7b

#15 = 7b

#16-21 = 77−7b

#22 = 7b

#23-28 = 77−7b

#29 = 7b

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

#39-44 = 77−b

#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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If $(2x+5)(x-3)=14$, find the sum of the possible values of $x$.
MA_775_DIABLO [31]

Answer:

x = 1/4 ± 1/4√233

Step-by-step explanation:

(2x + 5)(x - 3) = 14

~Use FOIL on the left side

2x² - 6x + 5x - 15 = 14

~Combine like terms

2x² - x - 15 = 14

~Subtract 14 to both sides

2x² - x - 29 = 0

~Use the quadratic formula and simplify

x = 1/4 ± 1/4√233

Best of Luck!

7 0
3 years ago
What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?
astraxan [27]
ANSWER


The value of the expression is
- 1


EXPLANATION

Method 1: Rewrite as product of
{i}^{2}


The expression given to us is,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}


We use the fact that
{i}^{2}  =  - 1
to simplify the above expression.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


We substitute to obtain,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  1 \times   1  \times  - 1 =  - 1


Method 2: Use indices to solve.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{0 + 1 + 2 + 3 + 4}



This implies that,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{10}




{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (  {{i}^{2}} )^{5}


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


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3 years ago
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abruzzese [7]

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Step-by-step explanation:

-13x = 25 [Divide by -13]

x = -1 12⁄13

13 is your <em>divisor</em>, which will be your denominator, and the numerator is your <em>remainder</em><em>.</em><em> </em>Since 13 goes into 25 ONCE, 1 gets multiplied by 13 to get 13, then deduct that from 25 to get your remainder of 12. This is where you start putting the pieces together as a mixed number. Then attach the negative because whenever you divide a positive by a negative, you get a negative:

-25⁄13 = -1 12⁄13

I am joyous to assist you anytime.

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Step-by-step explanation:

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