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Gnoma [55]
3 years ago
11

A grasshopper sits on the first square of a 1×N board. He can jump over one or two squares and land on the next square. The gras

shopper can jump forward or back but he must stay on the board. Find the least number n such that for any N ≥ n the grasshopper can land on each square exactly once. 100 PTS PLS HELP. Can someone answer this genuinely and not just take the points?!!? Thank you
Mathematics
2 answers:
Neko [114]3 years ago
6 0

Answer:

n=N-1

Step-by-step explanation:

Helga [31]3 years ago
4 0

Answer:

n=N-1

Step-by-step explanation:

You can start by imagining this scenario on a small scale, say 5 squares.

Assuming it starts on the first square, the grasshopper can cover the full 5 squares in 2 ways; either it can jump one square at a time, or it can jump all the way to the end and then backtrack. If it jumps one square at a time, it will take 4 hops to cover all 5 squares. If it jumps two squares at a time and then backtracks, it will take 2 jumps to cover the full 5 squares and then 2 to cover the 2 it missed, which is also 4. It will always be one less than the total amount of squares, since it begins on the first square and must touch the rest exactly once. Therefore, the smallest amount n is N-1. Hope this helps!

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If DE ⊥ EF, then ∠DEF is a right angle, provide a counter example if false... help??
svp [43]

Answer: true

Step-by-step explanation:

6 0
3 years ago
Solve for x. Your answer must be simplified. x-24>_9
3241004551 [841]

Answer:

x\geq33

Step-by-step explanation:

x-24\geq9

x\geq33

5 0
3 years ago
A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the wid
Mariana [72]

Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

We assume

Width = x

Length = 2x

Height = h

where, length = 2 \times width

Base area = lb

= 2x^2

Side area = 2lh + 2bh

= 2(2x)h + 2(x)h

= 4xh + 2xh

Volume = 24 which is lbh = 24

h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}

Now, cost is

= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}

now we have to minimize C(x)

So, we need to compute the C'(x)

= 52x - \frac{648}{x^2}

C"(x)  = 52x - \frac{1,296}{x^3}

now for the critical points, we will solve the equation C'(x) = 0

= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}

C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0

So, x is a point of minima that is

= (\frac{648}{52} )^\frac{1}{3}

Now, Base material cost is

= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}

= 139.75

Side material cost is

= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3}  }

= 279.50

and finally

Total cost is

= 139.75 + 279.50

= 419.25

4 0
3 years ago
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alukav5142 [94]

a/q b/a=70 so ans will be 210

8 0
3 years ago
Write an iequality for -5(×+7)&lt;-10 representing the solution for x​
vovikov84 [41]

Answer:

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Step-by-step explanation:

1) open the brackets

2) take -35 to the other side with -10

where it turns to be -5X<-10+35

3) work out to find -5X<25

4) divide both sides by -5

to find X<-5

3 0
3 years ago
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